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On: 20 October 2014, At: 02:54Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41Mortimer Street,

On: 20 October 2014, At: 02:54

Publisher: Taylor & Francis

Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41Mortimer Street, London W1T 3JH, UK

Journal of Difference Equations and Applications

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Published online: 17 Sep 2010

To cite this article: L.C Loi , N.H Du & P.K Anh (2002) On Linear Implicit Non-autonomous Systems of Difference Equations, Journal

of Difference Equations and Applications, 8:12, 1085-1105, DOI: 10.1080/1023619021000053962

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On Linear Implicit Non-autonomous Systems of Difference Equations

L.C LOI, N.H DU and P.K ANH*

Faculty of Mathematics, Mechanics and Informatics, Vietnam NationalUniversity, 334 Nguyen Trai, Thanh Xuan, Hanoi, Vietnam

(Received 11 July 2000; Revised 10 April 2001; In final form 27 April 2001)

This paper deals with the solvability of initial-value problems (IVPs) and multipoint value problems (MPBVPs) for linear implicit non-autonomous systems of difference equations.

boundary-Keywords: Linear implicit difference equations (LIDEs); Index of LIDEs; Singular-value decompostions (SVDs); IVPs; MPBVPs

1991 Mathematics Subject Classifications: 39A11; 39A10

ISSN 1023-6198 print/ISSN 1563-5120 online q 2002 Taylor & Francis Ltd

*Corresponding author Tel.: +740-593-9479 E-mail: anhpk@vnu.edu.vn

2002 Vol 8 (12), pp 1085–1105

2 THE CONSTANT-RANK CASE

Throughout this section, we assume that rankAn; r (n [ N), where 0 ,

r , m: Consider a SVD of An:

An¼ UnSnVTnþ1 ð3Þwhere Sn is a diagonal matrix with singular values: snð1Þ$snð2Þ$ · · · $

sðrÞn 0 on the main diagonal, i.e Sn¼ diagðsnð1Þ; ;sðrÞn ; 0; ; 0Þ and

Un, Vn+1are orthonormal matrices, i.e:

UTnUn¼ UnUT

n ¼ VT nþ1Vnþ1¼ Vnþ1VTnþ1¼ I

Here I and Ik ðk – mÞ denote the m £ m and k £ k- identity matrices,respectively Set

ðiiÞ Pnþ1¼ G21

ðiiiÞ G21n BnQn¼ Vnþ1QVTn ð6ÞPnþ1G21n BnQn¼ 0; Qnþ1G21n BnQn¼ Vnþ1QVT

nþ1G21n Bn¼ 0 andusing Eq (6), we get Q~2n¼ QnVnVT

nþ1G21n BnQnVnVT

nþ1G21n Bn¼QnVnQVT G21Bn¼ VnQVT G21Bn¼ ~Qn: Lemma 1 is proved A

Lemma 2 Let An¼ UnSnVTnþ1¼ UnSnVT

nþ1be two SVDs of An Then(i) The matrices GnU Anþ BnVnQVT

nþ1 and GnU Anþ BnVnQ  VT

nþ1 aresimultaneously non-singular

(ii) If Gnis non-singular, then

VnQVTnþ1G21n ¼ VnQ VTnþ1G21n ð9Þand

~

PnG21n21¼ ~PnG21n21 ð10Þwhere P˜nis defined in Lemma 1

Proof (i) Denote by S¯n the subspace {z: BnVn VTnþ1z[ ImAn} andsuppose that Gn is non-singular For any x [ Sn> KerAn there exists

z [Rm

such that BnVn VTnþ1x ¼ Anz; and therefore, Qnþ1G21n BnVn VTnþ1x ¼

Qnþ1G21n Anz: Using Eq (5), we get Qnþ1G21n Anz ¼ Qnþ1Pnþ1z ¼ 0: Thismeans

Qnþ1G21n Bny ¼ 0 ð11Þwhere y U VnVTnþ1x: On the other hand, since x [ KerAn, there exists

z[Rm such that x ¼ Qnþ1z; hence y ¼ VnVTnþ1Qnþ1z¼ VnQ VTnþ1z:Further, An21y ¼ Un21Sn21VT

nVnQ  VT nþ1z¼ 0; therefore y [ KerAn21and y ¼ Qnh for some h[Rm: By virture of Eqs (7) and (11), we getQnþ1G21n BnQnh¼ Vnþ1QVT

nh¼ 0: From the last relation, we find that

Sn> KerAn ¼ {0}: Now the further proof of part (i) follows the same line

as the proofs given in Ref [2] Indeed, suppose Gnx ¼ 0 then

BnVn VTnþ1Qnþ1x ¼ 2Anx [ ImAn; therefore  Qnþ1x [ Sn: On the otherhand, Qnþ1x [ KerAn; hence Qnþ1x [ Sn> KerAn¼ {0}: It followsthat Anx ¼ 0; and therefore, x ¼ Qnþ1x ¼ 0: Thus Gn is non-singulareither

(ii) First we note that both Qn and Qn are projections onto KerAn21,therefore QnQn ¼ Qn; i.e VnQVTnVnQ  VTn ¼ VnQ VTn: From the last relation,

it follows that

Q ¼ VTVnQVTVnQ ð12Þ

Now we shall prove Eq (9) Instead of Q in Gn; substituting Eq (12), weobtain:

nþ1Qnþ1Pnþ1¼ 0: Further,

by virtue of Eq (6), we have VnQVT

nþ1ðG21

n BnQnÞ VnQ VTnþ1¼ðVnQVT

nÞ VnQ VTnþ1¼ QnQn Vn VT

nþ1¼ QnVn VTnþ1¼ VnQ VTnþ1:Thus VnQVT

nþ1G21n Gn ¼ VnQ VTnþ1; which means Eq (9)

Finally, to prove Eq (11), we first note that G21n21Gn21 ¼ G21

n21ðAn21þBn21Vn21Q  VTnÞ: From Eqs (5), (6) and (12), it follows that G21

Lemma 2 guarantees that the following definition does not depend on thechosen SVDs of An

Definition 1 The LIDE (Eq (1)) is said to be index-1 if:

i¼0G21n212iBn212i; MðnÞk ¼Qk

i¼0G21 n212iBn212ið0 # k # n 2 1Þ:

Proof Relation (13) follows directly from Eq (9) and the definitions of ~Pnand ^Pn: Further, using Eq (8), we get

~PnMðnÞk ¼ ~PnYk

i¼0

~Pn2iG21n212iBn212i ð15Þ

^

PnMðnÞn ¼ ^PnYk

i¼0

^Pn2iG21n212iBn212i ð16Þ

Applying Eq (13) and taking into account Eq (10), we have ^Pn2iG21n212i¼

~

Pn2iG21n212i¼ ~Pn2iG21n212i: Finally, combining the last relation withEqs (13), (15) and (16), we come to Eq (14), as was to be proved ANow we are ready to consider IVPs and BVPs for LIDEs

Theorem 1 Let LIDE (Eq (1)) be index-1 Then

Proof Multiplying both sides of Eq (17) from the left by Pnþ1G21n and

Qnþ1G21n ; respectively, and applying Eqs (5) – (7), we come to thefollowing system:

Further, multiplying both sides of Eq (21) from left by VnVTnþ1; we obtain

VnQVTnþ1G21n BnPnxnþ Qnxnþ VnQVTnþ1G21n qn¼ 0Denoting Pnxnby un, from the last relation, we find xn ¼ Pnxnþ Qnxn¼

ðI 2 VnQVT

nþ1G21n BnÞun2 VnQVTnþ1G21n qn: Thus,

xn¼ ~Pnun2 VnQVTnþ1G21n qn ð22ÞObserving that un is a solution of the explicit system of differenceequations

unþ1¼ Pnþ1G21

n Bnunþ Pnþ1G21

n qnu0¼ u0U P0x0

i¼0

Pn2iG21n212iBn212iPkþ1G21n qk

þ PnG21 n21qn21:

Using the last relation and Eq (22) and arguing as in the proof of theLemma 3, we come to formula (19) Now suppose, we are given two SVDs

The next problems we shall be concerned with are frequently referred to

as MPBVPs:

Aixiþ1¼ Bixiþ qi i ¼ 0; n 2 1 ð23Þ

Xn i¼0

are arbitrary vectors

Put X0U ~P0; XiU ~PiMðiÞi21ði ¼ 1; n 2 1Þ; XnU PnMðnÞn21 and D ¼Pn

i¼0CiXi: Obviously, {Xi}ni¼0 are fundamental solutions of Eq (23), i.e.AiXiþ1 ¼ BiXiði ¼ 0; n 2 1Þ: Let Ai¼ UiSiVT

Definition 2 A MPBVP for index-1 LIDE is said to be regular if thefollowing regularity condition is valid:

i¼0CiXiþ Cnð Xn2 XnÞ ¼ D þ Cnð Pn2 PnÞMðnÞn21: Now suppose that

Eq (25) is valid Let ðxT

0;jTÞT[ Kerð DjCnQnÞ then

Dx0þ CnQn j¼ 0 ð26Þ

Observing that Qn ¼ QnQn; from Eq (26), we have Dx0 þ Cnð Pn2PnÞMðnÞn21x0þ CnQnQn j¼ 0: Since An21Pn¼ An21¼ An21Pn; it follows

An21ð Pn2 PnÞMðnÞn21x0¼ 0: This means ð Pn2 PnÞMðnÞn21x0¼ Qnzfor some



PnMðnÞn21¼ PnMðnÞn21 ¼ Xn: This implies Dx0 ¼Pn

i¼0CiXix0 ¼Pn

i¼0CiXiP0x0¼ 0: From Eq (26), it follows that CnQnj¼ 0 Since

QnQn ¼ Qn and Qnðzþ QnjÞ ¼ 0; we get Qnj¼ 2Qnz; and therefore

CnQnz¼ 0: On the other hand, as Xix0 ¼ XiP0x0¼ 0 ði ¼ 0; nÞ; we findthat Dx0þ CnQnz¼ 0: This means ð xT0;zTÞT [ KerðDjCnQnÞ ¼ KerR;hence Qnj¼ X 2 Qnz¼ 0; therefore ð xT

0;jTÞT[ Ker R; where

Thus, the inclusion Ker ð DjCnQnÞ , Ker R is proved To show the converseinclusion, we observe that for arbitrary ðxT

0;jTÞT[ Ker R; there hold therelations P0x0¼ P0Px0¼ 0 and Qnj¼ 0: This implies that Xix0¼ Xix0¼XiP0x0¼ 0; therefore Dx0 ¼ 0: Further, since CnQn j¼ 0; it follows Dx0þ

Proof It suffices to prove that the regularity of MPBVP Eqs (23) and (24)

is equivalent to the fact that the corresponding homogenous MPBVP:

Aixiþ1¼ Bixiði ¼ 0; n 2 1Þ ð27Þ

Xn i¼0

has only trivial solutions Suppose first that problem (27), (28) has onlytrivial solution and let ðxT0;jTÞT [ KerðDjCnQnÞ: Putting x*

i U Xix0ði ¼0; n 2 1Þ and x*U Xnx0þ Qnj; we find that {x*

i}ni¼0 is a solution ofEqs (27) and (28) From the assumption, it follows x*

i ¼ 0 ði ¼ 0; nÞ:

In particular X0x0 ¼ 0 and Xnx0þ Qnj¼ 0: Since X0x0 ¼ ðI 2Q0V0VT1G210 B0Þx0¼ 0; or Q0V0VT1G210 B0x0¼ x0 we get P0x0 ¼ 0;consequently Xnx0¼ XnP0x0¼ 0: Thus, Qnj¼ 0; hence ð xT

0;jTÞT[KerR: This means KerðDjCnQnÞ , KerR: To prove the converse inclusion,let ð xT

0;jTÞT[ KerR: Then P0x0 ¼ 0; therefore Xix0¼ XiP0x0 ¼ 0ði ¼0; nÞ; hence Dx0¼ 0: On the other hand, j[ KerQn; consequently

ð xT0;jTÞT[ KerðDjCnQnÞ: Now suppose that MPBVP Eqs (23) and (24)

is regular and let {xi}ni¼0 be a solution of Eqs (27) and (28) Then thereexist x0;j[Rm

such that xi¼ Xix0; ði ¼ 0; n 2 1Þ and xn ¼ Xnx0þ Qnj:From condition (28), it follows D x0þ CnQnj¼ 0 or ð xT

0;jTÞT[KerðDjCnQnÞ: The regularity condition ensures that ð xT

0;jTÞT [ KerR;

or P0x0 ¼ 0 and Qnj¼ 0: Since Xix0 ¼ XiP0x0¼ 0 ði ¼ 0; nÞ it follows

xi¼ 0 ði ¼ 0; nÞ: Thus problem (27), (28) has only trivial solution

3 THE NON-CONSTANT-RANK CASE

We begin this section by considering a LIDE (Eq (1)) with nestingkernels

Let Qnbe linear projections onto KerAnand PnU I 2 Qn:

Lemma 5 Condition (29) is equivalent to the relations:

x0¼ ðI 2 Q0G210 B0Þx02 Q0G210 q0

xn¼ ðI 2 QnG21

n BnÞ{Yn21 i¼0Pn212iG21n212iBn212ix0

þXn22 k¼0

Yn222k

i¼0Pn212iG21n212iBn212iðPkG21

k qkþ QkjkÞ

þ Pn21G21

n21qn21þ Qn21jn21} 2 QnG21n qnðn $ 1Þ; ð31Þwherej [Rmði ¼ 0; n 2 1Þ are arbitrary vectors

Proof By arguing as in the proof of Theorem 1, we see that LIDE (Eq (1))

is equivalent to the system:

Pnxnþ12 PnG21n BnPnxn¼ PnG21

xn¼ ðI 2 QnG21

n BnÞPnxn2 QnG21n qn ð33ÞDenoting Pnxnby unand taking into account relation (30) in Lemma 5, from

Eq (32), we have

Pnunþ12 PnG21n Bnun¼ PnG21

n qnThe last equation has solutions of the form

unþ1¼ PnG21

n Bnunþ PnG21

n qnþ Qnjn ð34Þwhere jn[Rm

are arbitrary vectors From Eq (34), we easily deducethat

un¼Yn21 i¼0Pn212iG21n212iBn212iu0

þXn22 k¼0

Yn222k

i¼0Pn212iG21n212iBn212iðPkG21

In particular, when KerAnare constant, then rankAn; r; Pn; P; Qn ; Qand LIDE (Eq (1)) is index-1

According to relations (7), PiG21i BiQi21ji21¼ PG21

i BiQji21¼ 0 andsolution formula (31) is the same as Eq (19)

We end this section by considering a direct consequence of Theorem 3 Let

An¼ UnSnVT

nþ1be a SVD of An, whereSn¼ diagðsnð1Þ; ;sðrn Þ

n ; 0; ; 0Þand rnU rankAn:

{

Corollary Consider a LIDE with non-decreasing ranks: rankAnþ1$rankAn: Suppose that for all n [ N; the matrices Anþ BnVnQnVT

nþ1;where

Snþ UT

nBnVnQn are all non-singular Now Theorem 3 ensures that LIDE(Eq (35)) has solutions yn, therefore xn¼ Vnynwill be solutions of LIDE(Eq (1)) The proof of Corollary is complete A

4 EXAMPLE

We shall illustrate the main results of the preceding sections by using somesimple examples Because of limitation of space, we shall not give lengthydetails

Example 1 Consider Eq (1) with the following data:

!ð36Þ

In this case, rankAn¼ 1; An¼ UnSnVTnþ1;

Un ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

1 þ n2p

Vnþ1¼ 1ffiffiffi

2p

Pn¼12

Simple calculation shows that

0B

1CA;

Since Gnðn $ 0Þ are non-singular, Eq (1) with data (36) is an index-1LIDE Theorem 1 ensures the unique solvability of the corresponding IVP

Using formula (19), we find that the IVP (Eqs (17) and (18)) has a uniquesolution

x0 ¼ ð x0ð1Þ; 21ÞT; xn¼ ðn 2 1Þ!Xn22

k¼0

1k!ð2n2

1CC

1CC

C;

qi¼

1i

i2

0BBB

1CC

1CC

1CC

1CC

C; g¼

102

0BBB

1CC

Let Ai¼ UiSiVTiþ1 be a SVD of Ai, where

@

1CC

A; S0¼

1 0 0

0 ffiffiffi2

p0

0 0 0

0BB

@

1CC

0 21= ffiffiffi

2

p1= ffiffiffi2p

0BB

@

1CC

A;

U1 ¼

ð2 þ ffiffiffi3

pÞ=ð3 þ ffiffiffi

3

p

Þ ffiffiffi3

p

ð ffiffiffi3

p

2 1Þ=6 21= ffiffiffi

3p

ð21 þ ffiffiffi

3

pÞ=ð3 þ ffiffiffi

p

þ 1Þ=6 1= ffiffiffi

3p

0BB

@

1CC

A;

S1 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3 þ ffiffiffi3pp

0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3 2 ffiffiffi3pp

0

0BBB

1CC

C;

V2 ¼

1= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3 þ ffiffiffi3pp

ð1 þ ffiffiffi3

pÞ= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

6 þ 2 ffiffiffi3pp

02ð1 þ ffiffiffi

3

pÞ=2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3 þ ffiffiffi3pp

1= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

6 þ 2 ffiffiffi3pp

1= ffiffiffi2p

ð1 þ ffiffiffi3

pÞ=2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3 þ ffiffiffi3pp

21= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

6 þ 2 ffiffiffi3pp

1= ffiffiffi2p

0BBB

1CCC

@

1CA; V0U I; Qi¼ ViQVi

0 1= ffiffiffi

2

p

1= ffiffiffi2p

0B

@

1CA;

G1¼ A1þ B1V1QVT2 ¼

1 21 1

1 1 2 1 2

0B

@

1CA

Obviously, G0, G1, both are non-singular Thus LIDE with data (37) isindex-1 Now we study the regularity of problems (23) and (24) with data(37) and (38)

@

1CA;

@

1CA;

@

1CA

1CA;

1C

1CAand

1CCCCCC

Observing that KerðDjC2Q2Þ ¼ KerR; we come to the conclusion that thethree-point BVP is regular, hence, it has a unique solution Simple calc-ulation gives x0¼ ð13; 13; 0ÞT; x1¼ ð1; 22; 22ÞT; x2 ¼ ð21; 28; 25ÞT:

Example 3 Let us consider LIDE (Eq (1)) with the data:

@

1CCCCA

0BBBB

@

1CCCCA

qn¼ ð1; 1; 1; 1ÞT; ðn $ 0Þ ð39Þ

Since KerA0¼ Span{ð0; 0; 0; 1ÞT; ð0; 0; 1; 0ÞT; ð0; 1; 0; 0ÞT}; KerA1¼Span{ð0; 0; 0; 1ÞT; ð0; 0; 1; 0ÞT}; KerAn¼ Span{ð0; 0; 0; 1ÞT}ðn $ 2Þ: Itfollows that KerA0 KerA1 KerA2¼ KerAnðn $ 3Þ:

@

1CCCCA

@

1CCCCA

@

1CCCCA

@

1CCCCA

@

1CCCCA

@

1CC

@

1CC

A;

qn¼

n121

0BB

@

1CC

The matrices Anhave SVDs: An¼ UnSnVT

nþ1; where

Un¼ 1ffiffiffi2p

0 2 ffiffiffi2

p0

0B

@

1CA; Sn¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2n2þ 2

@

1CA;

Vnþ1¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

n2þ 1p

@

1CA

Observing that rankA0¼ 1 , rankAn¼ 2 ðn $ 1Þ we get a LIDE with decreasing ranks Putting

@

1CA; QnU Q ¼

0 0 0

0 0 0

0 0 1

0B

@

1CA; Pn¼ I 2 Qn

ðn $ 0Þ; V0¼ I:

We find

~G0¼ A0þ B0V0Q0VT

@

1CAand

~

Gn¼ Anþ BnVnQnVT

nþ1¼

1 2 2nan n þ 2an 0nðn22 2nÞan ð2n 2 n2Þan n

1 þ nð2n 2 3Þan n 2 ð2n 2 3Þan 0

0B

@

1CA

where an U ðn22 2n þ 2Þ21=2ðn2þ 1Þ21=2: Since det ~G0¼ 21; det ~Gn¼nð2n 2 1Þðn2þ 1Þan–0 ðn $ 1Þ; the Corollary of Theorem 3 implies thesolvability of LIDE (Eq (1)) with data (40)