research article the solutions of second-order linear...

Hindawi Publishing CorporationDiscrete Dynamics in Nature and SocietyVolume 2013 Article ID 976914 4 pageshttpdxdoiorg1011552013976914

Research ArticleThe Solutions of Second-Order Linear MatrixEquations on Time Scales

Kefeng Li and Chao Zhang

School of Mathematical Sciences University of Jinan Jinan Shandong 250022 China

Correspondence should be addressed to Chao Zhang ss zhangcujneducn

Received 11 July 2013 Accepted 29 August 2013

Academic Editor Shurong Sun

Copyright copy 2013 K Li and C Zhang This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This paper studies the solutions of second-order linear matrix equations on time scales Firstly the necessary and sufficientconditions for the existence of a solution of characteristic equation are introduced then two diverse solutions of characteristicequation are applied to express general solution of the matrix equations on time scales

1 Introduction

In this paper we consider the solutions for the followingsecond-order linear matrix equations

[119875 (119905)119883Δ (119905)]Δ

+ 119876119883Δ (119905) + 119877 (119905)119883120590 (119905) = 0 119905 isin T (1)

where 119875(119905) = 119875120583(119905) 119877(119905) = 119877120583(119905) 119875119876 119877119883(119905) isin Rmtimesm119883Δ is the delta derivative 120590(119905) is the forward jump operators119883120590(119905) = 119883(120590(119905)) 120583(119905) is the graininess function and T is aninfinite isolated time scale which is given by

T = 1199050 1199051 1199052 119905

119899 (2)

As a tool for establishing a unified framework for contin-uous and discrete analysis a theory of dynamic equations onmeasure chains was introduced by Hilger in his PhD thesis[1] in 1988 In many cases it is necessary to study a specialcase of measure chains-time scales In the last decade theinvestigation of dynamic systems on time scales has involvedmuch interesting including quite a few fields such as thetheory of calculus the oscillation of the dynamic systemsthe eigenvalue problems and boundary value problems andpartial differential equations on time scales and so forth [2ndash5] Up to now there are few results about matrix equationson time scales In 1998 Agarwal and Bohner [6] studiedthe quadratic functionals for second order matrix equationson time scales in 2002 Erbe and Peterson [7] obtainedoscillation criteria for a second-order self-adjoint matrix dif-ferential equation on time scales in terms of the eigenvalues

of the coefficient matrices and the graininess function Thetheory of dynamic systems on time scales is of very importanttheoretical significance and has a wide range of applications

Based on the related literatures the researches of solu-tions of matrix difference or differential equations have fewresults In 1999 Barkatou [8] proposed an algorithm for therational solutions of special matrix difference equations anddiscussed their applications in 2003 Freiling and Hochhaus[9] investigated some properties of the solution for rationalmatrix difference equations in 2004 Xu and Zhang [10]studied the representation of general solution for secondorder homogeneous matrix difference equations in 2011Wu and Zhou [11] obtained the particular solutions of onekind of second order matrix differential equations Since thecontinuous case and the discrete case are two special cases oftime scales so we study the solutions of second order linearmatrix equations on time scales

This paper is organized as follows Section 2 introducessome basic concepts and fundamental theory about timescales By using the characteristic equation of the matrixequation the solutions of (1) are obtained which will be givenin Section 3

2 Preliminaries

In this section some basic concepts and some fundamentalresults on time scales are introduced

2 Discrete Dynamics in Nature and Society

Let T sub R be a nonempty closed subset Define theforward and backward jump operators 120590 120588 T rarr T by

120590 (119905) = inf 119904 isin T 119904 gt 119905 120588 (119905) = sup 119904 isin T 119904 lt 119905 (3)

where inf 0 = sup T sup 0 = inf T We put T119896 = T if T isunbounded above and T119896 = T (120588(max T)max T] otherwiseThe graininess functions ] 120583 T rarr [0infin) are defined by

120583 (119905) = 120590 (119905) minus 119905 ] (119905) = 119905 minus 120588 (119905) (4)

Let 119891 be a function defined on T 119891 is said to be (delta)differentiable at 119905 isin T119896 provided that there exists a constant 119886such that for any 120576 gt 0 there is a neighborhood 119880 of 119905 (ie119880 = (119905 minus 120575 119905 + 120575) cap T for some 120575 gt 0) with

1003816100381610038161003816119891 (120590 (119905)) minus 119891 (119904) minus 119886 (120590 (119905) minus 119904)1003816100381610038161003816 le 120576 |120590 (119905) minus 119904| forall119904 isin 119880

(5)

In this case denote119891Δ(119905) = 119886 If119891 is (delta) differentiable forevery 119905 isin T119896 then 119891 is said to be (delta) differentiable on T If 119891 is differentiable at 119905 isin T119896 then

119891Δ (119905) =

lim119904rarr 119905

119904isinT

119891 (119905) minus 119891 (119904)119905 minus 119904

if 120583 (119905) = 0

119891 (120590 (119905)) minus 119891 (119905)120583 (119905)

if 120583 (119905) gt 0

(6)

For convenience we introduce the following results ([3Lemma 1] and [4 Chapter 1]) which are useful in this paper

Lemma 1 Let 119891 119892 T rarr R and 119905 isin T119896 If 119891 and 119892 aredifferentiable at 119905 then 119891119892 is differentiable at 119905 and

(119891119892)Δ (119905) = 119891120590 (119905) 119892Δ (119905) + 119891Δ (119905) 119892 (119905)

= 119891Δ (119905) 119892120590 (119905) + 119891 (119905) 119892Δ (119905) (7)

3 Main Results

We assume throughout this paper that 119875 is an invertiblematrix It follows from Lemma 1 (6) and (2) that the matrixequation (1) can be written in the form

1198831205902

(119905) minus 119861119883120590 (119905) minus 119860119883 (119905) = 0 119905 isin T (8)

where 1198831205902

(119905) = 119883120590(120590(119905)) 119861 = 2119864119898minus 119875minus1119876 minus 119875minus1119877 119860 =

119875minus1119876 minus 119864119898 and 119864

119898is the identity matrix

Definition 2 The equation

119865 (Λ) = Λ2 minus 119861Λ minus 119860 = 0 (9)

where Λ isin Rmtimesm is called the characteristic equation of (8)

Definition 3 The functions

119865 (120582) = 1205822 minus 119861120582 minus 119860 ℎ (120582) = det119865 (120582) (10)

are called the eigenmatrix and eigenpolynomial of (8) Sucha 120582 which satisfies ℎ(120582) = 0 is called an eigenvalue of 119865(120582)

In order to study the solutions of matrix equation (8) wenow introduce some results about the characteristic equation(9)

Theorem 4 If 119860119861 = 119861119860 and there exists 119862 isin Rmtimesm such that1198622 = 1198612 + 4119860 119861119862 = 119862119861 then the characteristic equation (9)has solutions

Proof Let Λ1= (12)(119861 + 119862) Λ

2= (12)(119861 minus 119862) Then Λ

1

and Λ2are solutions of (9)

Remark 5 Generally if 119860119861 = 119861119860 and 1198622 = 1198612 + 4119860 wecannot easily get 119861119862 = 119862119861

Example 6 Let

119860 = (

120

1414

) 119861 = (1 02 minus1) (11)

Then

119862 = (radic3 01

1 + radic31) 119861119862 = (

radic3 05 + 2radic31 + radic3

minus1)

119862119861 = (radic3 0

3 + 2radic31 + radic3

minus1)

(12)

Obviously 119861119862 =119862119861

So Huang and Chen [12] and J Huang and H Huang [13]obtained the following results

Definition 7 Let 1205820be an eigenvalue of 119865(120582) Then

1198811205820

= 119909 | (12058220minus 1198611205820minus 119860) 119909 = 0 119909 isin Rmtimes1 (13)

is called a characteristic subspace of 119865(120582) corresponding to1205820 the nonzero vector 119909 of 119881

1205820

is called the eigenvectorcorresponding to 120582

0

Theorem8 There exist the diagonalizable solutions of charac-teristic equation (9) if and only if the dimension of the sum ofcharacteristic subspaces 119881

120582119894

is n that is dimsumℎ119894=1119881120582119894

= 119898where 120582

119894(119894 = 1 2 ℎ) are the distinct eigenvalues of 119865(120582)

Discrete Dynamics in Nature and Society 3

Definition 9 Let 120582119894 119894 = 1 2 119904 be eigenvalues of 119865(120582)

The extension vectors set of 119881120582119894

are as follows

119864119881120582119894

= (1199090 1199091 119909

119896) |

(1205822119894minus 119861120582119894minus 119860) 119909

0= 0

(1205822119894minus 119861120582119894minus 119860) 119909

1

= minus (2120582119894119864119898minus 119861) 119909

0 119896 le 119896

119894minus 1

(1205822119894minus 119861120582119894minus 119860) 119909

119896

= minus (2120582119894119864119898minus 119861) 119909

119896minus1minus 119909119896minus2119864119898

(14)

where 119896119894is the multiplicity of 120582

119894 (1199090 1199091 119909

119896) is called the

component of 119864119881120582119894

Theorem 10 The characteristic equation (9) has solu-tions if and only if there exist some components (119909

0 1199091

119909119896) (119910

0 1199101 119910

119902) of the set sum119904

119894=1119864119881120582119894

in (14) and theyare m linearly independent extension vectors

In the following we discuss the general solutions of thematrix equation (8) by using the solutions of the characteris-tic equation (9)

Theorem 11 If Λ1and Λ

2are two diverse solutions of the

characteristic equation (9) which satisfy Λ1Λ2= Λ2Λ1 and

Λ1minus Λ2is invertible then

119883(119905119899) = (

119899

sum119896=1

Λ119896minus11Λ119899minus1198962)119883(119905

1)

minus (119899minus1

sum119896=1

Λ119899minus1198961Λ1198962)119883(119905

0)

(15)

is the general solution of the matrix equation (8)

Proof Since Λ1and Λ

2are two solutions of (9) then

Λ21minus 119861Λ

1minus 119860 = 0

Λ22minus 119861Λ

2minus 119860 = 0

(16)

which deduce

(Λ1+ Λ2) (Λ1minus Λ2) = 119861 (Λ

1minus Λ2) (17)

Hence by the invertibility of Λ1minus Λ2 we have

119861 = Λ1+ Λ2 minus119860 = Λ

1Λ2

(Λ1minus Λ2)2 = 1198612 + 4119860

(18)

Let 119862 = Λ1minus Λ2 It follows from Λ

1Λ2= Λ2Λ1that

119860119861 = 119861119860 119860119862 = 119862119860 119861119862 = 119862119861 (19)

Replacing 119861 and 119860 with (12)(119861 + 119862) + (12)(119861 minus 119862) andminus(14)(119861 + 119862)(119861 minus 119862) in matrix equation (8) we obtain

21199091205902

(119905) minus (119861 minus 119862)119883120590 (119905) = 12(119861 + 119862)

times [2119883120590 (119905) minus (119861 minus 119862)119883 (119905)] (20)

Let

119884 (119905) = 2119883120590 (119905) minus (119861 minus 119862)119883 (119905) (21)

Then (20) can be rewritten as

119884120590 (119905) = 12(119861 + 119862)119884 (119905) (22)

Hence the general solution of (21) is as follows

119884 (119905119899) = 1

2119899(119861 + 119862)119899119884 (1199050) (23)

where 119884(1199050) = 2119883(119905

1) minus (119861 minus 119862)119883(119905

0) Putting 119884(119905

119899) into (21)

we can get the general solution of (8)

119909 (119905119899) = 1

2119899minus1(119861 minus 119862)119899minus1119883(1199051)

+119899minus1

sum119896=1

12119896(119861 minus 119862)119896minus1119884 (119905119899minus119896)

= minus 12119899119899minus1

sum119896=1

(119861 + 119862)119899minus119896(119861 minus 119862)119896119883(1199050)

+ 12119899minus1119899

sum119896=1

(119861 + 119862)119896minus1(119861 minus 119862)119899minus119896119883(1199051)

= (119899

sum119896=1

Λ119896minus11Λ119899minus1198962)119883(119905

1)

minus (119899minus1

sum119896=1

Λ119899minus1198961Λ1198962)119883(119905

0)

(24)



characteristic equation (9) and Λ1minus Λ2is invertible then

119883(119905119899) = Λ119899

11198621+ Λ11989921198622 (25)

where 1198621= (Λ2minus Λ1)minus1(Λ

2119883(1199050) minus 119883(119905

1)) and 119862

2= (Λ1minus

Λ2)minus1(Λ

1119883(1199050) minus 119883(119905

1)) is the general solution of the matrix

equation (8)

Proof Firstly we prove (25) is a solution of (8) Putting (25)into (8) we have

(Λ119899+211198621+ Λ119899+221198622) minus 119861 (Λ119899+1

11198621+ Λ119899+121198622)

minus 119860 (Λ11989911198621+ Λ11989921198622)

= (Λ21minus 119861Λ

1minus 119860)Λ119899

11198621+ (Λ22minus 119861Λ

2minus 119860)

times Λ11989921198622= 0

(26)


By using119883(1199050) = 1198621+1198622119883(1199051) = Λ

11198621+Λ21198622 andΛ

1minusΛ2

is invertible then we get

1198621= (Λ2minus Λ1)minus1 (Λ

2119883(1199050) minus 119883 (119905

1))


1119883(1199050) minus 119883 (119905

1))

(27)

Next we will prove all the solutions of (8) can be written intothe form of (25)

Obviously by 1198621= (Λ2minus Λ1)minus1(Λ

2119883(1199050) minus 119883(119905

1)) and

1198622= (Λ1minusΛ2)minus1(Λ

1119883(1199050)minus119883(119905

1)) we have119883(119905

0) = 1198621+1198622

119883(1199051)Λ11198621+ Λ21198622 then 119883(119905

0) and 119883(119905

1) are in the form of

(25) Assume the solution of (8) can be written in the form of(25) for 119899 le 119896 that is

119883(119905119896) = Λ119896

11198621+ Λ11989621198622

for 119899 le 119896 (28)

Then it follows from (8) and Λ1 Λ2being solutions of (9)

that

119909 (119905119896+3) = 119883120590

2

(119905119896+1) = 119861119883120590 (119905

119896+1) + 119860119883 (119905

119896+1)

= 119861119883 (119905119896+2) + 119860119883 (119905

119896+1)

= 119861 (Λ119896+211198621+ Λ119896+221198622) + 119860 (Λ119896+1

11198621+ Λ119896+121198622)

= (119861Λ1+ 119860)Λ119896+1

11198621+ (119861Λ

2+ 119860)Λ119896+1

21198622

= Λ21Λ119896+111198621+ Λ22Λ119896+121198622

= Λ119896+311198621+ Λ119896+321198622

(29)

By using the method of mathematical induction we have allthe solutions of (8) can be written in the form of (25)

Acknowledgments

This research was supported by the NNSF of China(Grant nos 11071143 and 11101241) the NNSF of ShandongProvince (Grant nos ZR2009AL003 ZR2010AL016 andZR2011AL007) and the Scientific Research and Develop-ment Project of Shandong Provincial Education Department(J11LA01)

References

[1] S Hilger Ein Ma szligkettenkalkul mit Anwendung auf Zentrums-mannigfaltigkeiten [PhD thesis] Universitat Wurzburg 1988

[2] R P Agarwal M Bohner and P J Y Wong ldquoSturm-Liouvilleeigenvalue problems on time scalesrdquo Applied Mathematics andComputation vol 99 no 2-3 pp 153ndash166 1999

[3] R P Agarwal andM Bohner ldquoBasic calculus on time scales andsome of its applicationsrdquo Results in Mathematics vol 35 no 1-2pp 3ndash22 1999

[4] M Bohner and A PetersonDynamic Equations on Time ScalesAn Introduction with Applications Birkhauser Boston MassUSA 2001

[5] M Bohner and A Peterson Advances in Dynamic Equations onTime Scales Birkhauser Boston Mass USA 2003

[6] R P Agarwal andM Bohner ldquoQuadratic functionals for secondorder matrix equations on time scalesrdquo Nonlinear AnalysisTheoryMethodsampApplications vol 33 no 7 pp 675ndash692 1998

[7] L Erbe and A Peterson ldquoOscillation criteria for second-order matrix dynamic equations on a time scalerdquo Journal ofComputational and Applied Mathematics vol 141 no 1-2 pp169ndash185 2002

[8] M A Barkatou ldquoRational solutions of matrix difference equa-tions the problem of equivalence and factorizationrdquo in Proceed-ings of the International Symposium on Symbolic and AlgebraicComputation vol 99 pp 277ndash282 1999

[9] G Freiling and A Hochhaus ldquoProperties of the solutions ofrational matrix difference equationsrdquo Computers amp Mathemat-ics with Applications vol 45 no 6ndash9 pp 1137ndash1154 2003

[10] J Xu and B Zhang ldquoThe solutions of second order homogenousmatrix difference equationsrdquo College Mathematics vol 20 pp96ndash101 2004

[11] Y Wu and D Zhou ldquoThe particular solutions to one kind ofsecond order matrix ordinary differential equationrdquo Journal ofFoshan University vol 29 pp 14ndash19 2011

[12] J Huang and J Chen ldquoThe diagonalizable solution of thequadratic matrix equation 1198601198832 + 119861119883 + 119862 = 0rdquo Mathematicsin Practice and Theory vol 37 pp 153ndash156 2007

[13] J Huang and H Huang ldquoThe solutions of second orderlinear matrix difference equations and its asymptotic stabilityrdquoMathematics in Practice and Theory vol 39 pp 250ndash254 2009

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Stochastic AnalysisInternational Journal of


Let T sub R be a nonempty closed subset Define theforward and backward jump operators 120590 120588 T rarr T by

120590 (119905) = inf 119904 isin T 119904 gt 119905 120588 (119905) = sup 119904 isin T 119904 lt 119905 (3)

where inf 0 = sup T sup 0 = inf T We put T119896 = T if T isunbounded above and T119896 = T (120588(max T)max T] otherwiseThe graininess functions ] 120583 T rarr [0infin) are defined by

120583 (119905) = 120590 (119905) minus 119905 ] (119905) = 119905 minus 120588 (119905) (4)

Let 119891 be a function defined on T 119891 is said to be (delta)differentiable at 119905 isin T119896 provided that there exists a constant 119886such that for any 120576 gt 0 there is a neighborhood 119880 of 119905 (ie119880 = (119905 minus 120575 119905 + 120575) cap T for some 120575 gt 0) with

1003816100381610038161003816119891 (120590 (119905)) minus 119891 (119904) minus 119886 (120590 (119905) minus 119904)1003816100381610038161003816 le 120576 |120590 (119905) minus 119904| forall119904 isin 119880

(5)

In this case denote119891Δ(119905) = 119886 If119891 is (delta) differentiable forevery 119905 isin T119896 then 119891 is said to be (delta) differentiable on T If 119891 is differentiable at 119905 isin T119896 then

119891Δ (119905) =

lim119904rarr 119905

119904isinT

119891 (119905) minus 119891 (119904)119905 minus 119904

if 120583 (119905) = 0

119891 (120590 (119905)) minus 119891 (119905)120583 (119905)

if 120583 (119905) gt 0

(6)

For convenience we introduce the following results ([3Lemma 1] and [4 Chapter 1]) which are useful in this paper

Lemma 1 Let 119891 119892 T rarr R and 119905 isin T119896 If 119891 and 119892 aredifferentiable at 119905 then 119891119892 is differentiable at 119905 and

(119891119892)Δ (119905) = 119891120590 (119905) 119892Δ (119905) + 119891Δ (119905) 119892 (119905)

= 119891Δ (119905) 119892120590 (119905) + 119891 (119905) 119892Δ (119905) (7)

3 Main Results

We assume throughout this paper that 119875 is an invertiblematrix It follows from Lemma 1 (6) and (2) that the matrixequation (1) can be written in the form

1198831205902

(119905) minus 119861119883120590 (119905) minus 119860119883 (119905) = 0 119905 isin T (8)

where 1198831205902

(119905) = 119883120590(120590(119905)) 119861 = 2119864119898minus 119875minus1119876 minus 119875minus1119877 119860 =

119875minus1119876 minus 119864119898 and 119864

119898is the identity matrix

Definition 2 The equation

119865 (Λ) = Λ2 minus 119861Λ minus 119860 = 0 (9)

where Λ isin Rmtimesm is called the characteristic equation of (8)

Definition 3 The functions

119865 (120582) = 1205822 minus 119861120582 minus 119860 ℎ (120582) = det119865 (120582) (10)

are called the eigenmatrix and eigenpolynomial of (8) Sucha 120582 which satisfies ℎ(120582) = 0 is called an eigenvalue of 119865(120582)

In order to study the solutions of matrix equation (8) wenow introduce some results about the characteristic equation(9)

Theorem 4 If 119860119861 = 119861119860 and there exists 119862 isin Rmtimesm such that1198622 = 1198612 + 4119860 119861119862 = 119862119861 then the characteristic equation (9)has solutions

Proof Let Λ1= (12)(119861 + 119862) Λ

2= (12)(119861 minus 119862) Then Λ

1

and Λ2are solutions of (9)

Remark 5 Generally if 119860119861 = 119861119860 and 1198622 = 1198612 + 4119860 wecannot easily get 119861119862 = 119862119861

Example 6 Let

119860 = (

120

1414

) 119861 = (1 02 minus1) (11)

Then

119862 = (radic3 01

1 + radic31) 119861119862 = (

radic3 05 + 2radic31 + radic3

minus1)

119862119861 = (radic3 0

3 + 2radic31 + radic3

minus1)

(12)

Obviously 119861119862 =119862119861

So Huang and Chen [12] and J Huang and H Huang [13]obtained the following results

Definition 7 Let 1205820be an eigenvalue of 119865(120582) Then

1198811205820

= 119909 | (12058220minus 1198611205820minus 119860) 119909 = 0 119909 isin Rmtimes1 (13)

is called a characteristic subspace of 119865(120582) corresponding to1205820 the nonzero vector 119909 of 119881

1205820

is called the eigenvectorcorresponding to 120582

0

Theorem8 There exist the diagonalizable solutions of charac-teristic equation (9) if and only if the dimension of the sum ofcharacteristic subspaces 119881

120582119894

is n that is dimsumℎ119894=1119881120582119894

= 119898where 120582

119894(119894 = 1 2 ℎ) are the distinct eigenvalues of 119865(120582)




are as follows

119864119881120582119894

= (1199090 1199091 119909

119896) |

(1205822119894minus 119861120582119894minus 119860) 119909

0= 0

(1205822119894minus 119861120582119894minus 119860) 119909

1

= minus (2120582119894119864119898minus 119861) 119909

0 119896 le 119896

119894minus 1

(1205822119894minus 119861120582119894minus 119860) 119909

119896

= minus (2120582119894119864119898minus 119861) 119909


(14)


119894 (1199090 1199091 119909


component of 119864119881120582119894


0 1199091

119909119896) (119910

0 1199101 119910

119902) of the set sum119904

119894=1119864119881120582119894







119883(119905119899) = (

119899

sum119896=1

Λ119896minus11Λ119899minus1198962)119883(119905

1)

minus (119899minus1

sum119896=1

Λ119899minus1198961Λ1198962)119883(119905

0)

(15)




Λ21minus 119861Λ

1minus 119860 = 0

Λ22minus 119861Λ

2minus 119860 = 0

(16)

which deduce

(Λ1+ Λ2) (Λ1minus Λ2) = 119861 (Λ

1minus Λ2) (17)


119861 = Λ1+ Λ2 minus119860 = Λ

1Λ2

(Λ1minus Λ2)2 = 1198612 + 4119860

(18)


1Λ2= Λ2Λ1that

119860119861 = 119861119860 119860119862 = 119862119860 119861119862 = 119862119861 (19)


21199091205902

(119905) minus (119861 minus 119862)119883120590 (119905) = 12(119861 + 119862)


Let

119884 (119905) = 2119883120590 (119905) minus (119861 minus 119862)119883 (119905) (21)


119884120590 (119905) = 12(119861 + 119862)119884 (119905) (22)


119884 (119905119899) = 1

2119899(119861 + 119862)119899119884 (1199050) (23)

where 119884(1199050) = 2119883(119905

1) minus (119861 minus 119862)119883(119905

0) Putting 119884(119905

119899) into (21)


119909 (119905119899) = 1


+119899minus1

sum119896=1


= minus 12119899119899minus1

sum119896=1

(119861 + 119862)119899minus119896(119861 minus 119862)119896119883(1199050)

+ 12119899minus1119899

sum119896=1


= (119899

sum119896=1

Λ119896minus11Λ119899minus1198962)119883(119905

1)

minus (119899minus1

sum119896=1

Λ119899minus1198961Λ1198962)119883(119905

0)

(24)




119883(119905119899) = Λ119899

11198621+ Λ11989921198622 (25)


2119883(1199050) minus 119883(119905

1)) and 119862

2= (Λ1minus

Λ2)minus1(Λ

1119883(1199050) minus 119883(119905


equation (8)


(Λ119899+211198621+ Λ119899+221198622) minus 119861 (Λ119899+1

11198621+ Λ119899+121198622)

minus 119860 (Λ11989911198621+ Λ11989921198622)

= (Λ21minus 119861Λ

1minus 119860)Λ119899

11198621+ (Λ22minus 119861Λ

2minus 119860)

times Λ11989921198622= 0

(26)


By using119883(1199050) = 1198621+1198622119883(1199051) = Λ

11198621+Λ21198622 andΛ

1minusΛ2



2119883(1199050) minus 119883 (119905

1))


1119883(1199050) minus 119883 (119905

1))

(27)



2119883(1199050) minus 119883(119905

1)) and


1119883(1199050)minus119883(119905

1)) we have119883(119905

0) = 1198621+1198622

119883(1199051)Λ11198621+ Λ21198622 then 119883(119905

0) and 119883(119905



119883(119905119896) = Λ119896

11198621+ Λ11989621198622

for 119899 le 119896 (28)


that

119909 (119905119896+3) = 119883120590

2

(119905119896+1) = 119861119883120590 (119905

119896+1) + 119860119883 (119905

119896+1)

= 119861119883 (119905119896+2) + 119860119883 (119905

119896+1)

= 119861 (Λ119896+211198621+ Λ119896+221198622) + 119860 (Λ119896+1

11198621+ Λ119896+121198622)

= (119861Λ1+ 119860)Λ119896+1

11198621+ (119861Λ

2+ 119860)Λ119896+1

21198622

= Λ21Λ119896+111198621+ Λ22Λ119896+121198622

= Λ119896+311198621+ Λ119896+321198622

(29)


Acknowledgments


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












are as follows

119864119881120582119894

= (1199090 1199091 119909

119896) |

(1205822119894minus 119861120582119894minus 119860) 119909

0= 0

(1205822119894minus 119861120582119894minus 119860) 119909

1

= minus (2120582119894119864119898minus 119861) 119909

0 119896 le 119896

119894minus 1

(1205822119894minus 119861120582119894minus 119860) 119909

119896

= minus (2120582119894119864119898minus 119861) 119909


(14)


119894 (1199090 1199091 119909


component of 119864119881120582119894


0 1199091

119909119896) (119910

0 1199101 119910

119902) of the set sum119904

119894=1119864119881120582119894







119883(119905119899) = (

119899

sum119896=1

Λ119896minus11Λ119899minus1198962)119883(119905

1)

minus (119899minus1

sum119896=1

Λ119899minus1198961Λ1198962)119883(119905

0)

(15)




Λ21minus 119861Λ

1minus 119860 = 0

Λ22minus 119861Λ

2minus 119860 = 0

(16)

which deduce

(Λ1+ Λ2) (Λ1minus Λ2) = 119861 (Λ

1minus Λ2) (17)


119861 = Λ1+ Λ2 minus119860 = Λ

1Λ2

(Λ1minus Λ2)2 = 1198612 + 4119860

(18)


1Λ2= Λ2Λ1that

119860119861 = 119861119860 119860119862 = 119862119860 119861119862 = 119862119861 (19)


21199091205902

(119905) minus (119861 minus 119862)119883120590 (119905) = 12(119861 + 119862)


Let

119884 (119905) = 2119883120590 (119905) minus (119861 minus 119862)119883 (119905) (21)


119884120590 (119905) = 12(119861 + 119862)119884 (119905) (22)


119884 (119905119899) = 1

2119899(119861 + 119862)119899119884 (1199050) (23)

where 119884(1199050) = 2119883(119905

1) minus (119861 minus 119862)119883(119905

0) Putting 119884(119905

119899) into (21)


119909 (119905119899) = 1


+119899minus1

sum119896=1


= minus 12119899119899minus1

sum119896=1

(119861 + 119862)119899minus119896(119861 minus 119862)119896119883(1199050)

+ 12119899minus1119899

sum119896=1


= (119899

sum119896=1

Λ119896minus11Λ119899minus1198962)119883(119905

1)

minus (119899minus1

sum119896=1

Λ119899minus1198961Λ1198962)119883(119905

0)

(24)




119883(119905119899) = Λ119899

11198621+ Λ11989921198622 (25)


2119883(1199050) minus 119883(119905

1)) and 119862

2= (Λ1minus

Λ2)minus1(Λ

1119883(1199050) minus 119883(119905


equation (8)


(Λ119899+211198621+ Λ119899+221198622) minus 119861 (Λ119899+1

11198621+ Λ119899+121198622)

minus 119860 (Λ11989911198621+ Λ11989921198622)

= (Λ21minus 119861Λ

1minus 119860)Λ119899

11198621+ (Λ22minus 119861Λ

2minus 119860)

times Λ11989921198622= 0

(26)


By using119883(1199050) = 1198621+1198622119883(1199051) = Λ

11198621+Λ21198622 andΛ

1minusΛ2



2119883(1199050) minus 119883 (119905

1))


1119883(1199050) minus 119883 (119905

1))

(27)



2119883(1199050) minus 119883(119905

1)) and


1119883(1199050)minus119883(119905

1)) we have119883(119905

0) = 1198621+1198622

119883(1199051)Λ11198621+ Λ21198622 then 119883(119905

0) and 119883(119905



119883(119905119896) = Λ119896

11198621+ Λ11989621198622

for 119899 le 119896 (28)


that

119909 (119905119896+3) = 119883120590

2

(119905119896+1) = 119861119883120590 (119905

119896+1) + 119860119883 (119905

119896+1)

= 119861119883 (119905119896+2) + 119860119883 (119905

119896+1)

= 119861 (Λ119896+211198621+ Λ119896+221198622) + 119860 (Λ119896+1

11198621+ Λ119896+121198622)

= (119861Λ1+ 119860)Λ119896+1

11198621+ (119861Λ

2+ 119860)Λ119896+1

21198622

= Λ21Λ119896+111198621+ Λ22Λ119896+121198622

= Λ119896+311198621+ Λ119896+321198622

(29)


Acknowledgments


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










By using119883(1199050) = 1198621+1198622119883(1199051) = Λ

11198621+Λ21198622 andΛ

1minusΛ2



2119883(1199050) minus 119883 (119905

1))


1119883(1199050) minus 119883 (119905

1))

(27)



2119883(1199050) minus 119883(119905

1)) and


1119883(1199050)minus119883(119905

1)) we have119883(119905

0) = 1198621+1198622

119883(1199051)Λ11198621+ Λ21198622 then 119883(119905

0) and 119883(119905



119883(119905119896) = Λ119896

11198621+ Λ11989621198622

for 119899 le 119896 (28)


that

119909 (119905119896+3) = 119883120590

2

(119905119896+1) = 119861119883120590 (119905

119896+1) + 119860119883 (119905

119896+1)

= 119861119883 (119905119896+2) + 119860119883 (119905

119896+1)

= 119861 (Λ119896+211198621+ Λ119896+221198622) + 119860 (Λ119896+1

11198621+ Λ119896+121198622)

= (119861Λ1+ 119860)Λ119896+1

11198621+ (119861Λ

2+ 119860)Λ119896+1

21198622

= Λ21Λ119896+111198621+ Λ22Λ119896+121198622

= Λ119896+311198621+ Λ119896+321198622

(29)


Acknowledgments


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article the solutions of second-order linear...

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