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...

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 1

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

1 Introduction

Linear singular difference equations (LSDEs) of the form

A k x k+1 = Bk x k + qk , k ∈ N,

whereN denotes either the set of natural numbers including zero or its subset, Ak , B k∈

Rm ×m , q

k∈ Rm are given, and the matrix A k 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 q k= 0

Further, we only consider the case whereN 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:

A k x k+1 = Bk x k , k = 0, , N − 1. (1)

L.C Loi (B)

Department of Mathematics, Vietnam National University, Hanoi, Vietnam

e-mail: loilc@vnu.edu.vn

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, R m ×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 bound-ary value problems [5], sensitivity analysis of DAEs [9,10], and control theory of sin-gular 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 ob-tained

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 2

we recall the notions used in the following sections In Sect.3we 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.4 Finally, in Sect.5we discuss some open problems

2 Basic Concepts

Adjoint or dual equations have been introduced in the theory of DAEs In [8], the dual equation of (2) has the form



A T (t )z

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

A T k−1 λ k−1= B T

k λ k , k = 1, , N − 1. (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 origi-nal 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 rankA k ≡ r, 1 ≤ r ≤ m − 1.

Definition 1 [2,3] Equation (1) is called index-1 tractable (index-1 for short) if the follow-ing conditions hold:

(i) rank A k = r, k = 0, , N − 1,

(ii) S k ∩ ker Ak−1= {0}, k = 1, , N − 1, where Sk = {z ∈ R m : 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 A N−1 ) = m − r.

Let Q k be a projector onto ker A k and put P k := I − Qk Since rank A k = rank Ak−1,

there exists T k ∈ GL(R m ) such that T k|ker A k is an isomorphism from ker A k onto ker A k−1

Such an operator can be constructed as follows Since Q k∈ Rm×m is a projector onto kerA

k,

there exists a nonsingular matrix Z k∈ Rm×msuch that

Q k = Zk QZ¯ k−1,

where ¯Q = diag(Or , I m−r ) and O r , I m−r stand for r × r zero and (m − r) × (m − r)

iden-tity matrices, respectively To simplify the notation we will again denote by T kthe matrix

induced by the operator T k It is easy to verify that

T k = Zk−1 Z−1k

In practice, to find any projector onto kerA kone can consider a singular value decomposition

(SVD) of the matrix A k,

where Σ k = diag(σ (1)

k , , σ k (r) , 0, , 0) with σ k (i) , i = 1, , r being singular values of

A k , and where U k , V kare orthogonal matrices Clearly,

is an orthogonal projector onto kerA k(see [3,15] for details) Moreover,

and T k|ker A k is an isomorphism from ker A k onto ker A k−1

In the following sections we focus on a subadjoint equation of (1); hence, in what follows,

for simplicity we will use orthogonal projectors Q k and operators T kgiven by (6) and (7), respectively

Together with (1), we introduce the following matrices:

G k := Ak + Bk T k Q k or G k := Ak + Bk V k−1 QV¯ T , k = 1, , N − 1.

Since the SVD (5) of the matrix Ak is not unique, the matrix G k depends on the chosen

SVDs of A k We need the following lemmas

Lemma 1 [2,3] The following assertions are equivalent:

(i) the matrix G k := Ak + Bk V k−1 QV¯ T

k is nonsingular;

(ii) S k ∩ ker Ak−1= {0};

(iii) Rm = Sk ⊕ ker Ak−1

Due to Lemma1, if (1) is of index 1, then the matrix Gkis nonsingular Moreover, its

nonsingularity does not depend on the choice of SVDs of A

For the index-1 LSDE (1), we have the decomposition

Rm = Sk ⊕ ker Ak−1, k = 1, , N − 1.

Let{s (1)

k , , s k (r) } and {h (r+1) k−1 , , h (m) k−1 } denote certain bases of Sk and kerA k−1, respec-tively In other words,

S k= spans k (i)r

i=1



, k = 0, , N − 1

and

ker A k−1= spanh (i) k−1m

i =r+1



, k = 1, , N.

Lemma 2 [2] Suppose (1) is of index 1 and Ak = Uk Σ k V T is an arbitrary SVD of A k

Then, the following relations hold:

(i) ˜Q k−1:= Vk−1QV¯ T G−1k B k is the canonical projector onto ker A k−1along S k for every

k = 1, , N − 1;

(ii) ˜Q k−1 = ˜Zk−1 Q ˜¯Z−1k−1, where ˜ Z k−1 is a matrix, whose columns are vectors s k (1) , , s k (r) and h (r k−1+1) , , h (m) k−1for every k = 1, , N − 1.

3 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 E k , k = 0, , N − 1 and a transformation of variables

x k = Fk−1 ¯xk , k = 0, , N, the index-1 LSDE (1) can be transformed into the following equation:

¯

A k ¯xk+1= ¯Bk ¯xk , k = 0, , N − 1,

where ¯A k = Ek A k F kand ¯B k = Ek B k F k−1 , k = 0, , N − 1 Moreover, both matrices Ek

and F k are nonsingular for all k The above equation is said to be in Kronecker normal form

if

¯

A k = diag(Ir , O m−r ), ¯B k = diag(Wk , I m−r ), k = 0, , N − 1.

The following theorem was proved

Theorem 1 [2] Suppose (1) is of index 1 Then there exist nonsingular scaling matrices

E k and transforming variable matrices F k such that (1) is transformed into the Kronecker

normal form

diag(I r , O m−r ) ¯xk+1 = diag(Wk , I m−r ) ¯xk , k = 0, , N − 1. (8)

Remark 1 From the proof of Theorem1(see [2, Theorem 3.3]), the scaling matrices Ek

and the transforming variable matrices F kfor the index-1 equation (1) can be constructed as follows Due to Lemmas1and2, the matrix

˜G k := Ak + Bk ˜Z k−1Q ˜¯Z−1

is nonsingular, and so is the matrix

¯G k := ˜Gk ˜Z k = Ak ˜Z k + Bk ˜Z k−1 Q.¯

Thus, we can use the scaling E k = ¯G−1

k and the transformation of variables F k = ˜Zk

As discussed in Remark1, in order to perform the scaling matrices Ek , k = 0, , N − 1

and the transforming variable matrices F k , k = −1, , N − 1 we must specify the matrices

¯G k := Ak ˜Z k + Bk ˜Z k−1 Q, k¯ = 0, , N − 1 and ˜Zk for k = −1, , N − 1 Assume that

families of vectors{h (i)

−1}m

i =r+1and{s (i)

N }r

i=1are linearly independent inRm, such that

S0∩ spanh (i)−1m

i=r+1



= {0}, span

s N (i)r

i=1



∩ ker AN−1= {0}.

With these assumptions, we obtain all necessary scaling equations and transforming vari-ables, reducing the index-1 equation (1) to the Kronecker normal form (8)

Remark 2 For (1) and (8), it is easy to check that

¯S k∩ ker ¯A k−1= F−1

k−1 (S k ∩ ker Ak−1), k = 1, , N − 1.

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

¯xk:=



¯x (1) k

¯x (2) k



, where¯x (1)

k ∈ Rrand ¯x (2)

k ∈ Rm −r .

Obviously, (8) is equivalent to the system



¯x (1)

k+1= Wk ¯x (1)

k ,

0= ¯x (2)

k ,

k = 0, , N − 1. (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

¯u (1)

k = W T

k ¯u (1) k+1 , k = 0, , N − 1. (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:

0= ¯u (2) k+1 , k = 0, , N − 1. (11) Equations (10) and (11) can be rewritten in a compact form:

diag(I r , O m−r ) T ¯uk = diag(Wk , I m−r ) T ¯uk+1 , k = 0, , N − 1, (12) where ¯uk := ( ¯u (1)T

, ¯u (2)T

) T

In the following, to simplify the notation we will write H −T := (H−1) T = (H T )−1,

where H∈ Rm×mis an invertible matrix Next, performing a transformation of variables

¯uk := E −T k−1 u k , k = 1, , N

for a subequation of (12), we obtain

diag(I r , O m−r ) T E k−1 −T u k = diag(Wk , I m−r ) T E −T k u k+1 , k = 1, , N − 1.

Multiplying the above equation by F k−1 −T gives



E k−1−1 diag(I r , O m−r )F k−1−1T

u k=E k−1diag(W k , I m−r )F k−1−1T

u k+1 ,

or equivalently,

A T k−1 u k = B T

k u k+1, k = 1, , N − 1. (13) 3.2 Inherent Regular Ordinary Difference Equation

We briefly describe the decomposition technique for index-1 LSDEs (see [2,3,13,15] for details) Multiplying (1) by Pk G−1k and Q k G−1k , respectively, and carrying out a few techni-cal computations, we decouple this equation into the system



P k x k+1 = Pk G−1k B k P k−1 x k ,

0= Vk−1 QV¯ k T G−1k B k P k−1 x k + Qk−1 x k ,

k = 1, , N − 1.

Decomposing the solution{xk} N

k=1of (1) into two parts,

x k = Pk−1 x k + Qk−1 x k =: yk + zk , k = 1, , N − 1,

we can rewrite the above system as

y k+1= Pk G−1k B k y k , k = 1, , N − 1 (14) and

z k = −Vk−1 QV¯ T G−1k B k y k , k = 1, , N − 1.

Clearly, we only need to initialise the P−1-component of x0, i.e.,

P−1

x0− x0

where x0is an arbitrary vector inRm It yields that the initial value problem (IVP) for the index-1 LSDE (1) and the initial condition (15) has a unique solution

x k= ˜P k−1y k , k = 1, , N − 1, (16) where{yk} N−1

k=1 solves the IVP for (14) and

y0= y0:= P−1 x0.

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)

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

w k=P k G−1k B k

T

w k+1 , k = 1, , N − 1,

or

w k = B T

k G −T k P k T w k+1 , k = 1, , N − 1. (17) Putting

v k:=A†k−1T

P T

−1w k , k = 1, , N,

where A†k−1 denotes the Moore–Penrose inverse of A k−1, and noting that

A†k−1 A k−1 = Vk−1 Σ k−1† U k−1 T U k−1 Σ k−1 V k−1 T

= Vk−1 diag(I r , O m−r )V k−1 T

= I − Vk−1 QV¯ k−1 T = Pk−1 ,

we find that

A T

−1v k=A†k−1A k−1T

P T

−1w k = P T

−1w k , k = 1, , N.

Multiplying (17) by PT

−1and using the above equalities, we get

A T k−1 v k=A k G−1k B k P k−1T

v k+1, k = 1, , N − 1. (18) Observing that

G k P k=A k + Bk V k−1 QV¯ T

k



P k = Ak (I − Qk ) + Bk V k−1 V k T Q k P k = Ak ,

one has

G−1k A k = Pk , k = 1, , N − 1.

This leads to the relations

A k G−1k B k Q k−1 = Ak G−1k B k V k−1 QV¯ T Q k V k V T−1

= Ak G−1k (G k − Ak )Q k V k V k−1 T = 0

and

A T k−1 G −T k−1 A T k−1=A k−1 G−1k−1 A k−1T

= (Ak−1 P k−1 ) T = A T

k−1 .

It follows that (18) can be rewritten as

A T k−1

G −T k−1 A T k−1 v k



= B T k



G −T k A T k v k+1

, k = 1, , N − 1.

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 2 Equation (13) is called the subadjoint equation of the index-1 LSDE (1)

Remark 3 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 equa-tion (13) is smaller than that of the original equaequa-tion (1) For this reason, the term ‘subad-joint’ rather than ‘ad‘subad-joint’ 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

4 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:

S k b=z∈ Rm : B T

k z ∈ im A T

k−1



, k = 1, , N − 1.

Definition 3 Equation (13) is called index-1 tractable (index-1 for short) if the following conditions hold:

(i) rank A T

−1= r, k = 1, , N − 1,

(ii) S b ∩ ker A T = {0}, k = 1, , N − 2.

We recall that

¯

A k := diag(Ir , O m −r ) = Ek A k F k , ¯B k := diag(Wk , I m −r ) = Ek B k F k−1,

where E k = ¯G−1

k and F k = ˜Zk; hence, it yields

A T k = F −T

k A¯T

k E k −T (k = 0, , N − 2), B k T = F −T

k−1 ¯B T

k E −T k (k = 1, , N − 1).

Further, note that

ker ¯A T=z T , z TT

∈ Rm : z1∈ Rr and z1= 0

and

¯S b

k:=z∈ Rm : ¯B T

k z∈ im ¯A T k−1

=z1T , z T2T

∈ Rm : z2∈ Rm −r and z

2= 0.

It is also easy to see that

ker A T k = E T

k ker ¯A T k (k = 0, , N − 2), S b k = E T

k ¯S b

k (k = 1, , N − 1).

This means that

S b ∩ ker A T = E T

ker ¯A T ∩ ¯S b

= E T {0}= {0} for all k = 1, , N − 2.

On the other hand, clearly

rank A T k−1= rank Ak−1= r, k = 1, , N − 1.

Thus, we have completed the proof of the following useful result

Theorem 2 Let the original equation (1) be of index 1 Then, its subadjoint equation (13)

is also of index 1.

Example 1 Let us consider equation (1) with the data

A k=

⎛

⎝k+ 11 00 k+ 11

⎞

⎠ , B k=

⎛

⎝1k k+ 21 k+ 21

⎞

⎠ (19)

Observing that rank A k = 2 ∀k = 0, , N − 1,

S k= span(1, 0, 1) T ,

0, 1, (k + 1)2+ kT, k = 0, , N − 1

and

ker A k−1= span(1, 0, −1) T

, k = 1, , N.

Hence, for any k = 1, , N − 1 we have

S k ∩ ker Ak−1= {0}.

On the other hand, by direct computations, we find that rank A T

−1= 2 for all k = 1, ,

N− 1,

ker A T k = span( −k − 1, 1, 0) T

, k = 0, , N − 2,

and

S b k= span(1, 0, 0) T , (0, −k + 1, 2) T

, k = 1, , N − 1.

It follows that

S k b ∩ ker A T

k = {0}, k = 1, , N − 2.

Thus, we come to the conclusion that (1) with the data (19) and its subadjoint equation are

of the same index 1

Example 2 This example taken from [13] shows that if the original equation is not of in-dex 1, then its subadjoint is not an inin-dex-1 equation Let

A k=

⎛

⎝1k 01 00

⎞

⎠ , B k=

⎛

⎝ −k12 −1 10 k

−k − 1 1 0

⎞

⎠ , k = 0, ,N − 1. (20)

Since ker A k−1 = span{(0, 0, 1) T } for all k = 1, , N and

S k= span(0, 0, 1) T , (1, k + 1, 0) T

, k = 0, , N − 1,

it yields that

S k ∩ ker Ak−1= span(0, 0, 1) T

, k = 1, , N − 1.

Thus, (1) with the data (20) is not an index-1 equation We now consider its subadjoint equation Noting that

ker A T= span(0, 0, 1) T

, k = 0, , N − 2

S k b= span(0, 0, 1) T , ( −k, 1, 0) T

, k = 1, , N − 1,

we have

S k b ∩ ker A T

k = span(0, 0, 1) T

, k = 1, , N − 2.

Therefore, neither (1) with the data (20) nor its subadjoint equation are of index 1

Since rank A T = r for all k = 0, , N − 1, i.e., dim(ker A T

−1) = dim(ker A T )for all

k = 1, , N − 1, we can define Tb,k−1∈ GL(R m ) such that T b,k−1|ker A T

−1is an isomorphism

from kerA T

−1onto ker A T Let Q b−1be any projector onto ker A T−1and P b−1:= I − Q b

−1.

To characterise equation (13) we need the notation

G b,k−1 := A T

k−1 + B T

k T b,k−1 Q b k−1 , k = 1, , N − 1.

Remark 4 In the index-1 linear DAE case, the relation between Q k and Q bhas been shown

in [6, Remark 1] Since both Qk and Q b are arbitrary, the equality Q T = Q bmay not hold

For example, consider the orthogonal projector P k := A†

k A k and P b := (A T )†A T or,

equiv-alently, P k = Vk P V¯ T and P b = Uk P U¯ T with ¯P := I − ¯ Q Therefore, the necessary and sufficient condition for Q T = Q b is that V k = Ukholds Obviously, this is not valid for an

arbitrary matrix A k ; however, it is true for symmetric matrices A k

According to the SVD (5), Qb

k−1 := Uk−1 QU¯ T

−1is an orthogonal projector onto ker A T−1

and T b,k−1| ker A T

−1 is an isomorphism from ker A T

−1 onto ker A T , where T b,k−1 := Uk U T

−1.

Similarly as in Sect.2, in the following we will use the orthogonal projectors Qb

k−1and the

operators T b,k−1 Therefore,

G b,k−1 = A T

−1+ B T U k QU¯ T

−1, k = 1, , N − 1.

Now we come to the following lemmas from the well-known results with A = A T

k−1 , ¯ A = A T

and B = B T (see [13, Lemmas A.1 and A.2]) and Theorem2

Lemma 3 The following conditions are equivalent:

(i) the matrix G b,k−1 := A T

k−1 + B T U k QU¯ T

k−1 is nonsingular;

(ii) Rm = S b ⊕ ker A T;

(iii) S b

k ∩ ker A T

k = {0}

Lemma 4 Suppose (1) is of index 1 Then, the following relations hold:

P k−1 b = G−1

G−1b,k−1B T = G−1

b,k−1B T P b + Uk−1 QU¯ T (22)

Further, ˜ Q b := Uk QU¯ k−1 T G−1b,k−1 B T is the canonical projector onto ker A T along S b Let ˜Q b = ˜Zb,k Q ˜¯Z b,k−1, ˜G b,k := A T + B T

+1˜Z b,k+1Q ˜¯Z−1

b,kand ˜P b := I − ˜ Q b The canonical projectors mentioned in Lemmas2and4corresponding 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]