numerical methods for a class of differential algebraic...

Research ArticleNumerical Methods for a Class of DifferentialAlgebraic Equations

Lei Ren and Yuan-MingWang

Department of Mathematics Shanghai Key Laboratory of Pure Mathematics and Mathematical PracticeEast China Normal University Shanghai 200241 China

Correspondence should be addressed to Lei Ren renleipost163com

Received 28 November 2016 Accepted 11 April 2017 Published 8 June 2017

Academic Editor Fazal M Mahomed

Copyright copy 2017 Lei Ren and Yuan-Ming Wang This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

This paper is devoted to the study of some efficient numerical methods for the differential algebraic equations (DAEs) At firstwe propose a finite algorithm to compute the Drazin inverse of the time varying DAEs Numerical experiments are presented byDrazin inverse and Radau IIA method which illustrate that the precision of the Drazin inverse method is higher than the RadauIIA method Then Drazin inverse Radau IIA and Pade approximation are applied to the constant coefficient DAEs respectivelyNumerical results demonstrate that the Pade approximation is powerful for solving constant coefficient DAEs

1 Introduction

The general differential algebraic equations (DAEs)119865 (119905 119909 (119905) 1199091015840 (119905)) = 0 (1)

arise in many applications circuit analysis control theorychemical process simulations singular perturbation con-strained mechanical systems of rigid bodies and so forth [1ndash3] In [4ndash6] the numerical solutions of constant coefficientDAEs were studied by Golub et al The linear time varying ofDAE 119860 (119905) 1199091015840 (119905) + 119861 (119905) 119909 (119905) = 119891 (119905) (2)

defined on the interval 119868 is important in understandinggeneral DAEs It exhibits most of the behavior found in thenonlinear case that is not already present in the constantcoefficient case yet the linearity facilitates the analysis Inspite of this many aspects of the theory and numericalsolution of (2) have only been resolved within the last fewyears Some questions remain open

In [7] one canonical form for higher index linear timevarying singular systems has been presented In [8ndash10]with the DAEs being tractable with indexes 2 and 3 someresults were proposed It was shown that either 2 or index 3

tractability causes the homogeneous equations with respectto (2) to provide a finite dimensional space of solutionsIn [11] Song had treated general higher index time varyinglinear DAEs in terms of matrix pencils and investigated theirbasic properties and their solvability

Above these results the numerical solutions of DAEsseldom are involved In this paper we shall consider thehomogeneous equations with respect to DAEs119860 (119905) 1199091015840 (119905) + 119861 (119905) 119909 (119905) = 0 (3)

In this paper we shall consider the constant coefficientand time varying DAEs with different numerical methodsIn Section 2 we present some definitions which will be usedin the proofs of our main theorems In Section 3 a finitealgorithm for the computation of the Drazin inverse of thetime varying singularmatrix119860(119905) is presentedDrazin inverseis applied to (3) corresponding to the difference equationswe will establish sufficient and necessary conditions forthe solution of (3) In Section 4 the numerical treatmentwith the corresponding difference format is presented thenthe numerical solutions of constant coefficient DAEs arepresented by Pade approximation and the implicit Runge-Kutta method In Section 5 some numerical examples anderror estimates are proposed

HindawiMathematical Problems in EngineeringVolume 2017 Article ID 1871590 10 pageshttpsdoiorg10115520171871590

2 Mathematical Problems in Engineering

2 Preliminaries

We first introduce some definitions of the Drazin inverse of amatrix

Definition 1 Let 119860 isin C119899times119899 If the smallest positive integer 119896such that

rank (119860119896+1) = rank (119860119896) (4)

holds it is called the index of119860 and is denoted by 119896 = Ind(119860)If A is nonsingular then Ind(119860) = 0 else if119860 is singular thenInd(119860) ge 1Definition 2 Let 119860 isin C119899times119899 and Ind(119860) = 119896 Then the matrix119883 isin C119899times119899 satisfying

(1119896) 119860119896119883119860 = 119860119896 (5)

(2) 119883119860119883 = 119883 (6)

(3) 119860119883 = 119883119860 (7)

is called the Drazin inverse of 119860 and is denoted by119883 = 119860119889Consider the following singular differential and corre-

sponding difference equations119860 (119905) 1199091015840 + 119861 (119905) 119909 (119905) = 0119860 (119905) 119909119899+1 = 119861 (119905) 119909119899 (8)

Suppose 119905119899+1 = 119905119899 + ℎ ℎ is sufficient small Then119860 (119905) 119909 (119905119899+1) minus 119909 (119905119899)119905119899+1 minus 119905119899 asymp 119860 (119905) 119909 (119905119899) = minus119861 (119905) 119909 (119905119899) (9)

Therefore the approximation of 1198601199091015840 is given by119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 minus ℎ119861 (119905) 119909119899 (10)

Let 119862(119905) = (119860(119905) minus ℎ119861(119905)) we can get the difference equations119860(119905)119909119899+1 = 119862(119905)119909119899 like (8) When 119860(119905) is a singular matrixthings can happen that are impossible when 119860(119905)minus1 existsDefinition 3 For 119860(119905) 119861(119905) isin C119899times119899(119905) and 1199050 isin 119877 the vector119888 isin C119899 is called a consistent initial vector associated with 1199050for the equation 119860(119905)1199091015840(119905) + 119861(119905)119909(119905) = 119891(119905) when the initialproblem 119860(119905)1199091015840(119905) + 119861(119905)119909(119905) = 119891(119905) 119909(1199050) = 119888 119899 = 1 2 processes at least one solution

Definition 4 The equation 119860(119905)1199091015840(119905) + 119861(119905)119909(119905) = 119891(119905) issaid to tractable at the point 1199050 if the initial value problem119860(119905)1199091015840(119905) + 119861(119905)119909(119905) = 119891(119905) 119909(1199050) = 119888 has a unique solutionfor each consistent initial vector 119888 associated with 1199050

Definition 5 For119860(119905) 119861(119905) isin C119899times119899(119905) and 119891119899 isin C119899 the vector119888 isin C119899 is called a consistent initial vector for the differenceequation 119860(119905)119909119899+1 = 119861(119905)119909119899 + 119891119899 if the initial value problem119860(119905)119909119899+1 = 119861(119905)119909119899 +119891119899 1199090 = 119888 119899 = 1 2 has a solution for119909119899Definition 6 The difference equation 119860(119905)119909119899+1 = 119861(119905)119909119899 + 119891119899is said to be tractable if the initial value problem 119860(119905)119909119899+1 =119861(119905)119909119899+119891119899 1199090 = 119888 119899 = 1 2 has a unique solution for eachconsistent initial vector 1198883 Time Varying DifferentialAlgebraic Equations

Lemma 7 (see [12]) For 119860(119905) isin 119877(119905)119899times119899 Ind(119860(119905)) = 119896 and119892 (120582 119905) = det (120582119868 minus 119860119896+1 (119905))= 1198920 (119905) 120582119899 + 1198921 (119905) 120582119899minus1 + sdot sdot sdot + 119892119899minus1 (119905) 120582+ 119892119899 (119905) (11)

is characteristic polynomial of 119860119896+1(119905) where 1198920(119905) = 1 Then119892119903+1(119905) = sdot sdot sdot = 119892119899(119905) = 0 119892119903(119905) = 0 and119860119889 (119905) = minus119892minus1119903 (119905) 119860119896 (119905) [(119860119896+1 (119905))119903minus1+ 1198921 (119905) (119860119896+1 (119905))119903minus2 + sdot sdot sdot + 119892119903minus2 (119905) (119860119896+1 (119905))+ 119892119903minus1 (119905) 119868119899] (12)

where 119903 = rank(119860(119905))Next the algorithm for the computation of the Drazin

inverse of a polynomial matrix is presented as follows

Step 1 119865119894(119905) and 119892119894(119905) are determined by the recursiverelationship as follows119865119894+1 (119905) = 119860119896+1 (119905) 119865119894 (119905) + 119892119894 (t) 119868119892119894+1 (119905) = minus tr (119860119896+1 (119905) 119865119894+1 (119905))(119894 + 1) (13)

and initial conditions are1198651 (119905) = 1198681198991198921 (119905) = minustr (119860119896+1 (119905)) (14)

Step 2 From rank(119860119896(119905)) = rank(119860119896+1(119905)) = 119903 we can seethat 119892119903+1(119905) = sdot sdot sdot = 119892119899(119905) = 0 119892119903(119905) = 0 Then119860119889 (119905) = minus119892minus1119903 (119905) 119860119896 (119905) [(119860119896+1 (119905))119903minus1+ 1198921 (119905) (119860119896+1 (119905))119903minus2 + sdot sdot sdot + 119892119903minus2 (119905) (119860119896+1 (119905))+ 119892119903minus1 (119905) 119868119899] = minus119892minus1119903 (119905) 119860119896 (119905) 119865119903 (119905) (15)

The algorithm could be performed by the symbol computa-tion package of Matlab

Mathematical Problems in Engineering 3

We are now ready to give solutions on correspondingsingular difference equation First of all in order to establisha sufficient and necessary condition the following lemma isimportant

Lemma 8 Let 119860(119905) 119861(119905) isin C119899times119899(119905) Assume that there exists120582 isin C such that (120582119860(119905) + 119861(119905))minus1 exists and let119860120582 (119905) = (120582119860 (119905) + 119861 (119905))minus1 119860 (119905) 119861120582 (119905) = (120582119860 (119905) + 119861 (119905))minus1 119861 (119905) (16)

Then 119860120582(119905)119861120582(119905) = 119861120582(119905)119860120582(119905)Proof If there exists 120582 isin C such that (120582119860(119905) + 119861(119905))minus1 existsthen120582119860120582 (119905) + 119861120582 (119905) = 120582 (120582119860 (119905) + 119861 (119905))minus1 119860 (119905)+ (120582119860 (119905) + 119861 (119905))minus1 119861 (119905)= (120582119860 (119905) + 119861 (119905))minus1 (120582119860 (119905) + 119861 (119905))= 119868 (17)

Thus119861120582(119905)119860120582(119905) = (119868 minus 120582119860120582(119905))119860120582(119905) = 119860120582(119905) minus 120582119860120582(119905)2 =119860120582(119905)(119868minus120582119860120582(119905)) = 119860120582(119905)119861120582(119905) Consequently119860120582(119905)119861120582(119905) =119861120582(119905)119860120582(119905)Theorem 9 For119860(119905) 119861(119905) isin C119899times119899(119905) the homogeneous differ-ence equation 119860 (119905) 119909119899+1 = 119861 (119905) 119909119899 (18)

is tractable if and only if there exists 120582 isin C such that (120582119860(119905) +119861(119905))minus1 existsProof We first prove the sufficiency Let 119860120582(119905) and 119861120582(119905) bedefined as those in (27) Clearly119860(119905)119909119899+1 = 119861(119905)119909119899 is tractableif and only if 119860120582(119905)119909119899+1 = 119861120582(119905)119909119899 is tractable

Since 120582119860120582(119905) + 119861120582(119905) = 119868 exist invertible matrix 119879(119905) isinC119899times119899(119905) so that119879minus1 (119905) 119860120582 (119905) 119879 (119905) = (119862 (119905) 00 119873 (119905))

119879minus1 (119905) 119861120582 (119905) 119879 (119905) = (119868 minus 120582119862 (119905) 00 119868 minus 120582119873 (119905))= (1198611 (119905) 00 1198612 (119905)) (19)

where 119862(119905) is invertible and 119873(119905) is nilpotent of index 119896 Let119909119899 = 119879(119905)119910119899(119905) Then the differential equation becomes(119862 (119905) 00 119873 (119905))(119910(1)119899+1 (119905)119910(2)119899+1 (119905))= (119868 minus 120582119862 (119905) 00 119868 minus 120582119873 (119905))(119910(1)119899 (119905)119910(2)119899 (119905)) (20)

or 119862 (119905) 119910(1)119899+1 (119905) = (119868 minus 120582119862 (119905)) 119910(1)119899 (119905) 119873 (119905) 119910(2)119899+1 (119905) = (119868 minus 120582119873 (119905)) 119910(2)119899 (119905) (21)

Since 119862(119905) is invertible 119862(119905)119910(1)119899+1(119905) = (119868 minus 120582119862(119905))119910(1)119899 (119905) istractable Thus it suffices to show that the second equation ofis tractable Let 119896 = Ind(119873(119905)) and multiply (32) by 119873119896minus1(119905)Then (119868 minus 120582119873(119905))119873119896minus1119910(2)119899 = 0 and hence 119873119896minus1119910(2)119899 = 0 So119873119896minus1119910(2)119899+1 = 0 Multiply (32) again by119873119896minus2 Then119873119896minus1119910(2)119899+1 =(119868 minus 120582119873(119905))119873119896minus2119910(2)119899 so 119873119896minus2119910(2)119899 = 0 Continuing in thismanner we get 119910(2)119899 (119905) = 0 and119873119910(2)119899+1 + (119868 minus 120582119873(119905))119910(2)119899 = 0 istrivially tractable

For the necessity assume that 119860(119905)119909119899+1 = 119861(119905)119909119899 istractable We need to show that there is a 120582 isin C such that120582119860(119905) + 119861(119905) is invertible Assume that this is not true Then120582119860(119905) + 119861(119905) is singular for all 120582 isin C This means that foreach 120582 isin C there is a vector V120582(119905) isin C(119905) such that (120582119860(119905) +119861(119905))V120582(119905) = 0 and V120582(119905) = 0 Let V1205821(119905) V1205822(119905) V120582119904(119905)be a finite linearly dependent set of such vectors Let 119909(120582119894)119899 =120582119899119894 V120582119894(119905) and let 1205721(119905) 1205722(119905) 120572119904(119905) ⫅ C(119905) be such thatsum119904119894=1 120572119894V120582119894(119905) = 0 where not all 120572119894 are 0Then 119911119899 = sum119904119894=1 120572119894119909(120582119894)119899is not identically zero and is easily seen to be a solution of(18) However 1199110 = sum119904119894=1 120582119894V120582119894(119905) = 0 Thus there are twodifferent solutions namely 119911119899 and 0 which satisfy the initialcondition 1199090 = 0 Therefore it is not tractable at 119899 = 0 whichcontradicts our hypothesis Hence (120582119860(119905) + 119861(119905))minus1 exists forsome 120582 isin C

The next lemma will be used to show that the solution ofthe difference equation is independent of the scalar 120582which isused in the expressions (120582119860(119905)+119861(119905))minus1 and (120582119860(119905)minus119861(119905))minus1Lemma 10 Assume that 119860(119905) 119861(119905) isin C119899times119899(119905) are such thatthere exists 120582 isin C such that (120582119860(119905)plusmn119861(119905))minus1 exists Let119860120582(119905) =(120582119860(119905) plusmn 119861(119905))minus1119860(119905) 119861120582(119905) = (120582119860(119905) plusmn 119861(119905))minus1119861(119905) For all120572 120573 isin C for which (120572119860(119905) plusmn119861(119905))minus1 and (120573119860(119905) plusmn119861(119905))minus1 existthe following statements are true119860119889120572 (119905) 119860120572 (119905) = 119860119889120573 (119905) 119860120573 (119905) 119860119889120572 (119905) 119861120572 (119905) = 119860119889120573 (119905) 119861120573 (119905) 119860120572 (119905) 119861119889120572 (119905) = 119860120573 (119905) 119860119889120573 (119905)

Ind (119860120572 (119905)) = Ind (119860120573 (119905)) 119877 (119860120572 (119905)) = 119877 (119860120573 (119905)) (22)

Theorem 11 If the homogeneous equation 119860(119905)119909119899+1 = 119861(119905)119909119899is tractable then the general solution is given by119909119899 = 119860 (119905) 119860119889 (119905) 119902 if 119899 = 0(119860119889 (119905) 119861 (119905))119899 119902 if 119899 = 1 2 119902 isin C

119899 (119905) (23)

where 119860(119905) = (120582119860(119905) minus 119861(119905))minus1119860(119905) and 119861(119905) = (120582119860(119905) minus119861(119905))minus1119861(119905) and 120582 isin C is such that (120582119860(119905) minus 119861(119905))minus1 exists


Furthermore 119888 isin C119899 is a consistent initial vector for if andonly if 119888 isin 119877(119860119896(119905)) = 119877(119860119889(119905)119860(119905)) where 119896 = Ind(119860(119905))In this case this unique solution subject to 1199090 = 119888 is given by119909119899 = (119860119889(119905)119861(119905))119899119888 119899 = 0 1 2 3 Proof Since (3) is tractable multiplying by (120582119860(119905) minus 119861(119905))minus1gives the equivalent equation 119860(119905)119909119899+1 = 119861(119905)119909119899 After asimilarity we get as in the proof of Theorem 9 that

(119862 (119905) 00 119873 (119905))(119910(1)119899+1 (119905)119910(2)119899+1 (119905))= (119868 + 120582119862 (119905) 00 119868 + 120582119873 (119905))(119910(1)119899 (119905)119910(2)119899 (119905))= (1198611 00 1198612)(119910(1)119899 (119905)119910(2)119899 (119905))

(24)

The difference equation is equivalent to the pair of equations119862 (119905) 119910(1)119899+1 (119905) = 1198611 (119905) 119910(1)119899 (119905) 119873 (119905) 119910(2)119899+1 (119905) = 1198612 (119905) 119910(2)119899 (119905) (25)

Since 1198612 is invertible the unique solution of the secondequation of (25) was 119910(2)119899+1(119905) = 119861minus1198962 (119905)119873119870119910(2)119899+119896 = 0 Butthe first equation of (25) is consistent for any 1199100(119905) and theunique solution is 119910(1)119899 (119905) = 119862minus119899(119905)(1198611(119905))119910(1)0 (119905) In terms ofthe original variables we have119909119899 = 119879 (119905) 119910119899 (119905) = 119879 (119905)

sdot (119862minus119899 (119905) (1198611 (119905))119899 00 0)119879minus1 (119905) 119879 (119905)sdot (119868 00 0)119879minus1 (119905) 119879 (119905) (119910(1)0 (119905)119910(2)0 (119905)) = (119860119889119861)119899 119902

(26)

where 119902 = (119910(1)0 (119905) 119910(2)0 (119905))119879 is arbitrary4 Linear Coefficient DifferentialAlgebraic Equations

In order to establish a sufficient and necessary condition forlinear coefficient DAEs the following lemma is important

Lemma 12 (see [13]) Let 119860 119861 isin C119899times119899 Assume that thereexists 120582 isin C such that (120582119860 + 119861)minus1 exists and let119860120582 = (120582119860 + 119861)minus1 119860119861120582 = (120582119860 + 119861)minus1 119861 (27)

Then 119860120582119861120582 = 119861120582119860120582

Theorem 13 For 119860 119861 isin C119899times119899 the homogeneous differenceequation 119860119909119899+1 = 119861119909119899 (28)

is tractable if and only if there exists 120582 isin C such that (120582119860+119861)minus1exists

Proof The proof follows fromTheorem 9

The next lemma will be used to show that the solution ofthe difference equation is independent of the scalar 120582 whichis used in the expressions (120582119860 + 119861)minus1 and (120582119860 minus 119861)minus1Lemma 14 Assume that119860 119861 isin C119899times119899 are such that there exists120582 isin C such that (120582119860 plusmn 119861)minus1 exists Let 119860120582 = (120582119860 plusmn 119861)minus1119860119861120582 = (120582119860 plusmn 119861)minus1119861 For all 120572 120573 isin C for which (120574119860 plusmn 119861)minus1 and(120575119860 plusmn 119861)minus1 exist the following statements are true119860119889120574119860120574 = 119860119889120575119860120575119860119889120574119861120574 = 119860119889120575119861120575119860120574119861119889120574 = 119860120575119860119889120575 (29)

At last we give the general solution of the singulardifference equation

Theorem 15 (see [14]) If the homogeneous equation 119860119909119899+1 =119861119909119899 is tractable then the general solution is given by

119909119899 = 119860119860119889119902 if 119899 = 0(119860119889119861)119899 119902 if 119899 = 1 2 119902 isin C119899 (30)

where 119860 = (120582119860 minus 119861)minus1119860 and 119861 = (120582119860 minus 119861)minus1119861 and 120582 isin C issuch that (120582119860minus119861)minus1 exists Furthermore 119888 isin C119899 is a consistentinitial vector if and only if 119888 isin 119877(119860119896) = 119877(119860119889(119905)119860) where119896 = Ind(119860) In this case this unique solution subject to 1199090 = 119888is given by 119909119899 = (119860119889119861)119899119888 119899 = 0 1 2 3 Proof The proof follows fromTheorem 11

41 Radau IIA Methods We will introduce the IRK whichwould be applied concisely IRK methods play an importantrole for the numerical solution of DAEs Due to their one-step nature IRK methods are potentially more efficient forthese problems than multistep methods because multistepmethods must be restarted usually at low order after everydiscontinuity whereas IRK methods can restart at a higherorder

The 119904-stage implicit Runge-Kutta method applied to thegeneral nonlinear DAE of the form (2) is defined by0 = 119865 (119905119894 119883119894 119870119894) (31)119905119894 = 119905119899minus1 + 119888119894ℎ 119894 = 1 2 119904 (32)


119883119894 = 119909119899minus1 + ℎ 119904sum119895=1

119886119894119895119870119895 (33)

119909119899 = 119909119899minus1 + ℎ 119904sum119894=1

119887119894119870119894 (34)

The method is often denoted by the shorthand notation orButcher diagram

119888 A119879 =1198881 11988611 11988611 sdot sdot sdot 11988611199041198882 11988621 11988622 sdot sdot sdot 1198862119904 d

119888119904 1198861199041 1198861199042 sdot sdot sdot 1198861199041199041198871 1198872 sdot sdot sdot 119887119904 (35)

42 Pade Approximation [15] A differential algebraic equa-tion has the form 119865 (119905 119909 ) = 0 (36)

with initial values 119909 (1199050) = 1199090 (1199050) = 1199091 (37)

where 119865 and 119909 are vector functions for which we assumedsufficient differentiability

We assume the solution has the form119909 = 1199090 + 1199091119905 + 1198901199052 (38)

where 119890 is a vector function which is the same size as 1199090 and11990910158400 Substituting (38) into (36) and neglecting higher orderterm we have the linear equation of 119890 in the form119860119890 = 119861 (39)

where 119860 and 119861 are constant matrixes Solving (39) thecoefficients of 1199052 can be determined Repeating the aboveprocedure for higher order terms we can get the arbitraryorder power series of the solutions for (2)

The power series given by above procedure can betransformed into Pade series and we have numerical solutionof differential algebraic equation into (2) the specific stepscan be followed by (31)ndash(37) in [15]

5 Numerical Experiments

Example 1 We consider the following differential algebraicequation 119860 + 119861119909 = 0 (40)

where

119860 = (2 3 21 0 minus20 0 0 ) 119861 = ( 18 14 100 1 2minus27 minus21 minus15) (41)

and we take ℎ = 01 the analytical solution of (40) is 119909(119905) =(16 + 1611989021199053 minus13 + 4311989021199053 16 minus 13611989021199053)119879(1) Drazin Inverse Suppose 119905119899+1 = 119905119899+ℎ ℎ is sufficiently smallThen 119860119909 (119905119899+1) minus 119909 (119905119899)119905119899+1 minus 119905119899 asymp 119860119909 (119905119899) = minus119861119909 (119905119899) (42)

so on the point 119905119899+1 = 119905119899 + ℎ the approximate value of 1198601199091015840 isgiven by 119860119909119899+1 = 119860119909119899 minus ℎ119861119909119899 (43)

Now we use the difference rule to solve the equation Set

119862 = 119860 minus ℎ119861 = (02000 16000 1000010000 minus01000 minus2200027000 21000 15000 ) (44)

and then

119860 + 119862 = (22000 46000 3000020000 minus01000 minus4200027000 21000 15000 ) (45)

det(119860 + 119862) = minus3348 = 0 so 120582 = 1 such that (120582119860 + 119862)minus1exists we can see the homogeneous equation 119860119909119899+1 = 119862119909119899 istractable For minus119860 minus 119862 = minus(119860 + 119862) we have 120582 = minus1 such that(120582119860 minus 119862)minus1 exists Thus

119860 = (minus119860 minus 119862)119860 = ( 05000 07769 05538minus10000 minus12849 minus0569905000 04005 minus01989) 119862 = (minus119860 minus 119862)119862 = (minus15000 minus07769 minus0553810000 02849 05699minus05000 minus04005 minus08011) (46)

The eigenvalues of 119860119889 are 0 minus05 and minus04839 so that 119860119889could be computed byTheorem 752 in [13]

119860119889 = ( 20000 31148 22296minus40000 minus50815 minus2163020000 15074 minus09852) (47)


According toTheorem 11 the general solution is given by119909119899=(minus10000 minus15556 minus1111120000 25556 11111minus10000 minus07778 04444 )119902 if 119899 = 0(minus10000 minus15593 minus111852000 25259 10519minus1000 minus07296 05407 )119899 119902 if 119899 = 1 2

(48)

where 119902 = 119909(0) = (13 1 minus2)119879 which is a consistent initialvector

(2) Radau IIA Methods We consider Radau IIA as follows13 512 minus 1121 34 1434 14(49)

The 2-stage implicit Runge-Kutta method is applied to (40)1198601198701198991 + 119861(119909119899 + ℎ ( 5121198701198991 minus 1121198701198992)) = 01198601198701198992 + 119861(119909119899 + ℎ (341198701198991 minus 141198701198992)) = 0 (50)

We can rewrite (50) as follows

(119860 + 512ℎ119861 minus 112ℎ11986134ℎ119861 119860 + 14ℎ119861)(11987011989911198701198992) = (minus119861119909119899minus119861119909119899) (51)

where the coefficient matrix is nonsingular So we cancompute the numerical solutions by the following equation119909119899+1 = 119909119899 + ℎ (341198701198991 + 141198701198992) (52)

(3) Pade Approximation From initial values 119909(0) =(13 1 minus2)119879 the solutions of (40) can be supposed as1199091 (119905) = 13 + 11989011199051199092 (119905) = 1 + 11989021199051199093 (119905) = minus2 + 1198903119905 (53)

Substituting (53) into (40) and neglecting higher order termswe have (21198901 + 31198902 + 21198903) + 119874 (119905) = 0(1198901 minus 21198903) + 119874 (119905) = 0(91198901 + 71198902 + 51198903) 119905 = 0 (54)

The linear equation can be given in the following

(2 3 21 0 minus29 7 5 )(119890111989021198903) = (000) (55)

Solving this linear equation we have 119890 = (19 89 minus139)119879then the solutions of (40) can be supposed as

1199091 (119905) = 13 + 19119905 + 11989011199051199092 (119905) = 1 + 89119905 + 11989021199051199093 (119905) = minus2 minus 139119905 + 1198903119905(56)

In the same manner substituting (56) into (40) and neglect-ing higher order terms then we have

(41198901 + 61198902 + 41198903) 119905 + 119874 (1199052) = 0(21198901 minus 41198903 minus 2) 119905 + 119874 (1199052) = 0(minus271198901 minus 211198902 minus 151198903) 1199052 = 0 (57)

Similar to (53) solving this linear equation we have 119890 =(127 827 minus1327)119879 then the solutions of (40) can besupposed as

1199091 (119905) = 13 + 19119905 + 1271199052 + 119890111990531199092 (119905) = 1 + 89119905 + 8271199052 + 119890211990531199093 (119905) = minus2 minus 139119905 minus 13271199052 + 11989031199053(58)

Repeating the above procedure we have

1199091 (119905) = 13 + 19119905 + 1271199052 + 22431199053 + 17291199054 + 2109351199055+ 2984151199056 + 420667151199057 + 162001451199058+ 21674039151199059 + 119874 (11990510)


1199092 (119905) = 1 + 89119905 + 8271199052 + 162431199053 + 87291199054 + 16109351199055+ 16984151199056 + 3220667151199057 + 862001451199058+ 161674039151199059 + 119874 (11990510) 1199093 (119905) = minus2 minus 139119905 minus 13271199052 minus 262431199053 minus 137291199054 minus 26109351199055

minus 26984151199056 minus 5220667151199057 minus 1362001451199058minus 261674039151199059 + 119874 (11990510) (59)

The power series can be transformed into the following Padeseries

V1 (119905) = [54] = 13 + 181119905 + 42431199052 + 117011199053 + 1153091199054 + 168890511990551 minus 827119905 + 1271199052 minus 417011199053 + 1153091199054 V2 (119905) = [54] = 1 + 1627119905 + 172431199052 + 4451031199053 + 19459271199054 + 868890511990551 minus 827119905 + 1271199052 minus 417011199053 + 1153091199054 V3 (119905) = [54] = minus2 minus 2327119905 minus 312431199052 minus 6751031199053 minus 32459271199054 minus 1368890511990551 minus 827119905 + 1271199052 minus 417011199053 + 1153091199054

(60)

If the power series 1199091(119905) 1199092(119905) 1199093(119905) converges very fastthen 1199091(119905) 1199092(119905) 1199093(119905) can be eliminated in Pade series

In Tables 1ndash5 exact solutions numerical solutions anderrors are illustrated

Example 2 (see [16]) We consider the following differentialalgebraic equation119860 (119905) 1199091015840 (119905) + 119861 (119905) 119909 (119905) = 0 (61)

where

119860 = (1 0 0 00 1 0 00 0 1 00 0 0 0)119861 = ( 1 0 minus119905 1minus1 1 minus1199052 119905minus1199053 1199052 minus1 0119905 minus1 119905 minus1)

(62)

and we take ℎ = 001 the analytical solution of (40) is 119909(119905) =(119890minus119905 119905119890minus119905 119890minus119905 119905119890119905)119879(1) Drazin Inverse The classical four-order Runge-Kuttamethod is applied to (40) we can get

119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 + ℎ [1198961 + 21198962 + 21198963 + 1198964]6 (63)

where 1198961 = minus119861 (119905) 1199091198991198962 = minus119861 (119905) + ℎ11989612 1198963 = minus119861 (119905) + ℎ11989622 1198964 = minus119861 (119905) + ℎ1198963(64)

Thus

119860 (119905) 119909119899+1 = [119860 (119905) minus (ℎ + ℎ22 + ℎ36 + ℎ424)119861 (119905)] 119909119899 (65)

Nowwe use the difference rule to solve the system of (40)Set

119862 (119905) = [119860 (119905) minus (ℎ + ℎ22 + ℎ36 + ℎ424)119861 (119905)]= ( 08948 0 01052119905 minus0105201052 08948 010521199052 minus01052119905010521199053 minus010521199052 11052 0minus01052119905 01052 minus01052119905 01052 ) (66)

set 120582 = minus1 we get det(minus119860(119905) minus119862(119905)) = 07925 = 0 so we have120582 = minus1 such that (120582119860(119905) minus 119862(119905))minus1 exists Thus


Table 1 Exact solution and numerical solution of 119909(1)(119899)119905 (ℎ = 01) 119909(1)(119899) 119909(1)119899 of Drazin inv 119909(1)119899 of Radau 119909(1)119899 of Pade approx01 03448231843 03443666667 03448231345 0344823184202 03571051353 03563563033 03571050289 0357105135403 03702337930 03691464551 03702336224 0370233793104 03842675287 03827904314 03842672855 0384267528905 03992687375 03973451205 03992684126 0399268737806 04153041163 04128712011 04153036995 0415304116607 04324449597 04294333945 04324444400 0432444959308 04507674776 04471007347 04507668426 0450767477209 04703531334 04659468548 04703523699 0470353133510 04912890068 04860502943 04912881000 04912890064


Table 3 Exact solution and numerical solution of 119909(3)(119899)119905 (ℎ = 01) 119909(3)(119899) 119909(3)119899 of Drazin inv 119909(3)119899 of Radau 119909(3)119899 of Pade approx01 minus21493680625 minus21443333333 minus21493674154 minus2149368062202 minus23090334256 minus22981718600 minus23090320423 minus2309033425403 minus24797059760 minus24622586591 minus24797037581 minus2479705976204 minus26621445395 minus26372765649 minus26621413784 minus2662144539805 minus28571602544 minus28239538938 minus28571560306 minus2857160254606 minus30656201782 minus30230674884 minus30656147604 minus3065620178707 minus32884511428 minus32354459509 minus32884443862 minus3288451142408 minus35266438749 minus34619730912 minus35266356207 minus3526643874509 minus37812574008 minus37035916057 minus37812474748 minus3781257400310 minus40534237556 minus39613070007 minus40534119663 minus40534237552

Table 4 Errors 119864119899 fl 119909(119899) minus 1199091198992119905 (ℎ = 01) 119864119899 of Drazin inverse 119864119899 of Radau 119864119899 of Pade approximation01 59575917383119890 minus 003 76143396932119890 minus 007 59160779268119890 minus 01002 12916314922119890 minus 002 16276920132119890 minus 006 45825760741119890 minus 01003 20777743067119890 minus 002 26097631713119890 minus 006 24494899455119890 minus 01004 29630445273119890 minus 002 37196628261119890 minus 006 53851633972119890 minus 01005 39572170567119890 minus 002 49700993038119890 minus 006 46904157928119890 minus 01006 50709583610119890 minus 002 63750949076119890 minus 006 65574377970119890 minus 01007 63158869695119890 minus 002 79504631349119890 minus 006 82462103179119890 minus 01008 77046497345119890 minus 002 97126908162119890 minus 006 75498350599119890 minus 01009 83510065024119890 minus 002 11679886007119890 minus 005 59160780206119890 minus 01010 91969922436119890 minus 002 13872451324119890 minus 005 64031244204119890 minus 010


Table 5 Errors of Drazin inverse and Radau IIA119905 (ℎ = 001) 119864119899 of Drazin inverse 119864119899 of Radau IIA001 11228248162119890 minus 004 10088260628119890 minus 003002 63035622437119890 minus 004 86145976589119890 minus 003003 55249099531119890 minus 004 12036909993119890 minus 003004 39265521061119890 minus 004 13035988488119890 minus 003005 17558562753119890 minus 004 55498031923119890 minus 003006 17313059801119890 minus 004 67699688079119890 minus 003007 12203796056119890 minus 004 71281533560119890 minus 003008 50551368477119890 minus 004 96806736561119890 minus 003009 66468312281119890 minus 004 99061602809119890 minus 00301 61145689285119890 minus 004 13892356616119890 minus 003

119860 (119905) = (minus119860 (119905) minus 119862 (119905)) 119860 = ( minus05294 minus 00294119905 00294 0 000294 minus 002941199052 minus05294 + 00294119905 0 0000151199052 + 002641199053 minus002641199052 minus0475 0minus00294 minus 05294119905 + 000151199053 + 002641199054 05294 minus 002641199053 minus0475119905 0)119862 (119905) = (minus119860 (119905) minus 119862 (119905)) 119862 = ( minus04975 + 00025119905 minus00025 0 0minus00025 + 000251199052 minus04975 minus 00025119905 0 0minus000251199053 000251199052 minus05025 000025 + 05025119905 minus 000251199054 minus05025 + 000251199053 04975119905 minus1)

(67)

so that 119860119889 could be computed by finite algorithm (13)ndash(15)

119860119889 = ( minus18948 + 01052119905 minus01052 0 0minus01052 + 010521199052 minus18948 minus 01052119905 0 0minus010521199053 010521199052 minus21052 001052 minus 18948119905 minus 010521199054 18948 + 010521199053 minus21052119905 0) (68)

According toTheorem 15 the general solution is given by

119909119899 =

((

1minus 28544119905 28544 0 028544 minus 285441199052 1 + 28544119905 0 014261199052 + 256921199053 minus256921199052 1 0minus28544 + 119905 + 14261199053 + 256921199054 minus1 minus 256921199053 119905 0))119902 if 119899 = 0

((

08948 minus 01052119905 01052 0 001052 minus 010521199052 01052119905 0 0minus18334119905 + 010521199053 minus010521199052 11052 0minus01052 + 08948119905 minus 183341199053 + 010521199054 minus08948 minus 010521199053 11052119905 0))119902 if 119899 = 1 2

(69)

where 119902 isin C119899(119905)


Let 119902 isin 119877(119860(119905))119896 we can compute the numerical solutionof (40) by (33)

(2) Radau IIA We consider the Radau IIA as follows13 512 minus 1121 34 1434 14(70)

Radau IIA is applied to (40)1198601198701198991 + 119861(119909119899 + ℎ ( 5121198701198991 minus 1121198701198992)) = 01198601198701198992 + 119861(119909119899 + ℎ (341198701198991 minus 141198701198992)) = 0 (71)




See Table 5 for errors of the numerical results Error 119864119899 isdefined by 119864119899 fl 119909(119899) minus 11990911989926 Conclusion

The fundamental goal of this study has been to constructapproximations to numerical solutions of linear DAEs FromTable 4 we know that the numerical solutions of Padeapproximation approximate the exact solutions better thanother methods and the process of Pade approximation iseasily implemented Moreover Drazin inverse is appliedto solve the time varying differential algebraic equationsAccording to the obtained solutions we infer that Drazininverse is a powerful tool for solving this kind of problemsFromTable 5we know that the precision of theDrazin inversemethod is higher than the Radau IIA method but the Drazininverse method is implemented in more complex way thanthe Radau IIA method

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This work was supported by Science and TechnologyCommission of Shanghai Municipality (STCSM) (no13dz2260400)

References

[1] K E Brenan S L Campbell and L R PetzoldNumerical Solu-tion of Initial-Value Problems inDifferential-Algebraic EquationsNorth-Holland New York NY USA 1989

[2] S L Campbell Singular SystemofDifferential Equations PitmanPublishing Ltd San Francisco 1980

[3] S L Campbell Singular System of Differential Equations IIPitman Publishing Ltd San Francisco 1982

[4] GHGolub andC FVanMatrixComputations JohnsHopkinsUniversity Press Baltimore 1983

[5] B Kagstrom and A RuheMatrix Pencils Springer-Verlag 1983[6] J H Wilkinson ldquoThe practical significance of the drazin

inverserdquo in Recent Applications of Generalized Inverses S LCampbell Ed pp 82ndash99 Pitman 1982

[7] S L Campbell ldquoOne canonical form for higher-index lineartime-varying singular systemsrdquo Circuits Systems and SignalProcessing vol 2 no 3 pp 311ndash326 1983

[8] R Marz ldquoIndex-2 differential-algebraic equationsrdquo Results inMathematics vol 15 pp 149ndash171 1989

[9] R Marz ldquoSome new results concerning index-3 differential-algebraic equationsrdquo Journal of Mathematical Analysis andApplications vol 140 no 1 pp 177ndash199 1989

[10] R Marz ldquoHigher-index differential-algebraic equations anal-ysis and numerical treatmentrdquo Numerical Analysis vol 24 pp199ndash222 1990

[11] Y Z Sun ldquoSolvability of higher index time-varying lineardifferential-algebraic equationsrdquo Acta Mathematica Scientiavol B no 1 pp 77ndash92 2001

[12] J Gao and G Wang ldquoTwo algorithms of Drazin inverse of apolynomial matrixrdquo Joural of Shanghai Normal University vol6 (Science Edition) 2002

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman Great Britain 1979

[14] S L Campbell and C D Meyer Generalized Inverses of LinearTransformations Pitman Publishing Ltd London 2009

[15] C Ercan andBMustafa ldquoArbitrary order numericalmethod forsolving differential-algebraic equation by Pade seriesrdquo AppliedMathematics and Computation vol 137 pp 57ndash65 2003

[16] S Karimi Vanani and A Aminataei ldquoNumerical solution ofdifferential algebraic equations using a multiquadric approxi-mation schemerdquoMathematical andComputerModelling vol 53no 5-6 pp 659ndash666 2011

Submit your manuscripts athttpswwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Journal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


2 Preliminaries

We first introduce some definitions of the Drazin inverse of amatrix

Definition 1 Let 119860 isin C119899times119899 If the smallest positive integer 119896such that

rank (119860119896+1) = rank (119860119896) (4)

holds it is called the index of119860 and is denoted by 119896 = Ind(119860)If A is nonsingular then Ind(119860) = 0 else if119860 is singular thenInd(119860) ge 1Definition 2 Let 119860 isin C119899times119899 and Ind(119860) = 119896 Then the matrix119883 isin C119899times119899 satisfying

(1119896) 119860119896119883119860 = 119860119896 (5)

(2) 119883119860119883 = 119883 (6)

(3) 119860119883 = 119883119860 (7)

is called the Drazin inverse of 119860 and is denoted by119883 = 119860119889Consider the following singular differential and corre-

sponding difference equations119860 (119905) 1199091015840 + 119861 (119905) 119909 (119905) = 0119860 (119905) 119909119899+1 = 119861 (119905) 119909119899 (8)

Suppose 119905119899+1 = 119905119899 + ℎ ℎ is sufficient small Then119860 (119905) 119909 (119905119899+1) minus 119909 (119905119899)119905119899+1 minus 119905119899 asymp 119860 (119905) 119909 (119905119899) = minus119861 (119905) 119909 (119905119899) (9)

Therefore the approximation of 1198601199091015840 is given by119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 minus ℎ119861 (119905) 119909119899 (10)

Let 119862(119905) = (119860(119905) minus ℎ119861(119905)) we can get the difference equations119860(119905)119909119899+1 = 119862(119905)119909119899 like (8) When 119860(119905) is a singular matrixthings can happen that are impossible when 119860(119905)minus1 existsDefinition 3 For 119860(119905) 119861(119905) isin C119899times119899(119905) and 1199050 isin 119877 the vector119888 isin C119899 is called a consistent initial vector associated with 1199050for the equation 119860(119905)1199091015840(119905) + 119861(119905)119909(119905) = 119891(119905) when the initialproblem 119860(119905)1199091015840(119905) + 119861(119905)119909(119905) = 119891(119905) 119909(1199050) = 119888 119899 = 1 2 processes at least one solution

Definition 4 The equation 119860(119905)1199091015840(119905) + 119861(119905)119909(119905) = 119891(119905) issaid to tractable at the point 1199050 if the initial value problem119860(119905)1199091015840(119905) + 119861(119905)119909(119905) = 119891(119905) 119909(1199050) = 119888 has a unique solutionfor each consistent initial vector 119888 associated with 1199050

Definition 5 For119860(119905) 119861(119905) isin C119899times119899(119905) and 119891119899 isin C119899 the vector119888 isin C119899 is called a consistent initial vector for the differenceequation 119860(119905)119909119899+1 = 119861(119905)119909119899 + 119891119899 if the initial value problem119860(119905)119909119899+1 = 119861(119905)119909119899 +119891119899 1199090 = 119888 119899 = 1 2 has a solution for119909119899Definition 6 The difference equation 119860(119905)119909119899+1 = 119861(119905)119909119899 + 119891119899is said to be tractable if the initial value problem 119860(119905)119909119899+1 =119861(119905)119909119899+119891119899 1199090 = 119888 119899 = 1 2 has a unique solution for eachconsistent initial vector 1198883 Time Varying DifferentialAlgebraic Equations

Lemma 7 (see [12]) For 119860(119905) isin 119877(119905)119899times119899 Ind(119860(119905)) = 119896 and119892 (120582 119905) = det (120582119868 minus 119860119896+1 (119905))= 1198920 (119905) 120582119899 + 1198921 (119905) 120582119899minus1 + sdot sdot sdot + 119892119899minus1 (119905) 120582+ 119892119899 (119905) (11)

is characteristic polynomial of 119860119896+1(119905) where 1198920(119905) = 1 Then119892119903+1(119905) = sdot sdot sdot = 119892119899(119905) = 0 119892119903(119905) = 0 and119860119889 (119905) = minus119892minus1119903 (119905) 119860119896 (119905) [(119860119896+1 (119905))119903minus1+ 1198921 (119905) (119860119896+1 (119905))119903minus2 + sdot sdot sdot + 119892119903minus2 (119905) (119860119896+1 (119905))+ 119892119903minus1 (119905) 119868119899] (12)

where 119903 = rank(119860(119905))Next the algorithm for the computation of the Drazin

inverse of a polynomial matrix is presented as follows

Step 1 119865119894(119905) and 119892119894(119905) are determined by the recursiverelationship as follows119865119894+1 (119905) = 119860119896+1 (119905) 119865119894 (119905) + 119892119894 (t) 119868119892119894+1 (119905) = minus tr (119860119896+1 (119905) 119865119894+1 (119905))(119894 + 1) (13)

and initial conditions are1198651 (119905) = 1198681198991198921 (119905) = minustr (119860119896+1 (119905)) (14)

Step 2 From rank(119860119896(119905)) = rank(119860119896+1(119905)) = 119903 we can seethat 119892119903+1(119905) = sdot sdot sdot = 119892119899(119905) = 0 119892119903(119905) = 0 Then119860119889 (119905) = minus119892minus1119903 (119905) 119860119896 (119905) [(119860119896+1 (119905))119903minus1+ 1198921 (119905) (119860119896+1 (119905))119903minus2 + sdot sdot sdot + 119892119903minus2 (119905) (119860119896+1 (119905))+ 119892119903minus1 (119905) 119868119899] = minus119892minus1119903 (119905) 119860119896 (119905) 119865119903 (119905) (15)

The algorithm could be performed by the symbol computa-tion package of Matlab










or 119862 (119905) 119910(1)119899+1 (119905) = (119868 minus 120582119862 (119905)) 119910(1)119899 (119905) 119873 (119905) 119910(2)119899+1 (119905) = (119868 minus 120582119873 (119905)) 119910(2)119899 (119905) (21)




Ind (119860120572 (119905)) = Ind (119860120573 (119905)) 119877 (119860120572 (119905)) = 119877 (119860120573 (119905)) (22)


119899 (119905) (23)




(119862 (119905) 00 119873 (119905))(119910(1)119899+1 (119905)119910(2)119899+1 (119905))= (119868 + 120582119862 (119905) 00 119868 + 120582119873 (119905))(119910(1)119899 (119905)119910(2)119899 (119905))= (1198611 00 1198612)(119910(1)119899 (119905)119910(2)119899 (119905))

(24)




(26)




Then 119860120582119861120582 = 119861120582119860120582







119909119899 = 119860119860119889119902 if 119899 = 0(119860119889119861)119899 119902 if 119899 = 1 2 119902 isin C119899 (30)





119883119894 = 119909119899minus1 + ℎ 119904sum119895=1

119886119894119895119870119895 (33)

119909119899 = 119909119899minus1 + ℎ 119904sum119894=1

119887119894119870119894 (34)













where






and then

119860 + 119862 = (22000 46000 3000020000 minus01000 minus4200027000 21000 15000 ) (45)







(48)










(2 3 21 0 minus29 7 5 )(119890111989021198903) = (000) (55)


1199091 (119905) = 13 + 19119905 + 11989011199051199092 (119905) = 1 + 89119905 + 11989021199051199093 (119905) = minus2 minus 139119905 + 1198903119905(56)






1199091 (119905) = 13 + 19119905 + 1271199052 + 22431199053 + 17291199054 + 2109351199055+ 2984151199056 + 420667151199057 + 162001451199058+ 21674039151199059 + 119874 (11990510)






(60)




where


(62)


119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 + ℎ [1198961 + 21198962 + 21198963 + 1198964]6 (63)


Thus

119860 (119905) 119909119899+1 = [119860 (119905) minus (ℎ + ℎ22 + ℎ36 + ℎ424)119861 (119905)] 119909119899 (65)












(67)




119909119899 =

((


((


(69)

where 119902 isin C119899(119905)












Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of













or 119862 (119905) 119910(1)119899+1 (119905) = (119868 minus 120582119862 (119905)) 119910(1)119899 (119905) 119873 (119905) 119910(2)119899+1 (119905) = (119868 minus 120582119873 (119905)) 119910(2)119899 (119905) (21)




Ind (119860120572 (119905)) = Ind (119860120573 (119905)) 119877 (119860120572 (119905)) = 119877 (119860120573 (119905)) (22)


119899 (119905) (23)




(119862 (119905) 00 119873 (119905))(119910(1)119899+1 (119905)119910(2)119899+1 (119905))= (119868 + 120582119862 (119905) 00 119868 + 120582119873 (119905))(119910(1)119899 (119905)119910(2)119899 (119905))= (1198611 00 1198612)(119910(1)119899 (119905)119910(2)119899 (119905))

(24)




(26)




Then 119860120582119861120582 = 119861120582119860120582







119909119899 = 119860119860119889119902 if 119899 = 0(119860119889119861)119899 119902 if 119899 = 1 2 119902 isin C119899 (30)





119883119894 = 119909119899minus1 + ℎ 119904sum119895=1

119886119894119895119870119895 (33)

119909119899 = 119909119899minus1 + ℎ 119904sum119894=1

119887119894119870119894 (34)













where






and then

119860 + 119862 = (22000 46000 3000020000 minus01000 minus4200027000 21000 15000 ) (45)







(48)










(2 3 21 0 minus29 7 5 )(119890111989021198903) = (000) (55)


1199091 (119905) = 13 + 19119905 + 11989011199051199092 (119905) = 1 + 89119905 + 11989021199051199093 (119905) = minus2 minus 139119905 + 1198903119905(56)






1199091 (119905) = 13 + 19119905 + 1271199052 + 22431199053 + 17291199054 + 2109351199055+ 2984151199056 + 420667151199057 + 162001451199058+ 21674039151199059 + 119874 (11990510)






(60)




where


(62)


119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 + ℎ [1198961 + 21198962 + 21198963 + 1198964]6 (63)


Thus

119860 (119905) 119909119899+1 = [119860 (119905) minus (ℎ + ℎ22 + ℎ36 + ℎ424)119861 (119905)] 119909119899 (65)












(67)




119909119899 =

((


((


(69)

where 119902 isin C119899(119905)












Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of






(119862 (119905) 00 119873 (119905))(119910(1)119899+1 (119905)119910(2)119899+1 (119905))= (119868 + 120582119862 (119905) 00 119868 + 120582119873 (119905))(119910(1)119899 (119905)119910(2)119899 (119905))= (1198611 00 1198612)(119910(1)119899 (119905)119910(2)119899 (119905))

(24)




(26)




Then 119860120582119861120582 = 119861120582119860120582







119909119899 = 119860119860119889119902 if 119899 = 0(119860119889119861)119899 119902 if 119899 = 1 2 119902 isin C119899 (30)





119883119894 = 119909119899minus1 + ℎ 119904sum119895=1

119886119894119895119870119895 (33)

119909119899 = 119909119899minus1 + ℎ 119904sum119894=1

119887119894119870119894 (34)













where






and then

119860 + 119862 = (22000 46000 3000020000 minus01000 minus4200027000 21000 15000 ) (45)







(48)










(2 3 21 0 minus29 7 5 )(119890111989021198903) = (000) (55)


1199091 (119905) = 13 + 19119905 + 11989011199051199092 (119905) = 1 + 89119905 + 11989021199051199093 (119905) = minus2 minus 139119905 + 1198903119905(56)






1199091 (119905) = 13 + 19119905 + 1271199052 + 22431199053 + 17291199054 + 2109351199055+ 2984151199056 + 420667151199057 + 162001451199058+ 21674039151199059 + 119874 (11990510)






(60)




where


(62)


119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 + ℎ [1198961 + 21198962 + 21198963 + 1198964]6 (63)


Thus

119860 (119905) 119909119899+1 = [119860 (119905) minus (ℎ + ℎ22 + ℎ36 + ℎ424)119861 (119905)] 119909119899 (65)












(67)




119909119899 =

((


((


(69)

where 119902 isin C119899(119905)












Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





119883119894 = 119909119899minus1 + ℎ 119904sum119895=1

119886119894119895119870119895 (33)

119909119899 = 119909119899minus1 + ℎ 119904sum119894=1

119887119894119870119894 (34)













where






and then

119860 + 119862 = (22000 46000 3000020000 minus01000 minus4200027000 21000 15000 ) (45)







(48)










(2 3 21 0 minus29 7 5 )(119890111989021198903) = (000) (55)


1199091 (119905) = 13 + 19119905 + 11989011199051199092 (119905) = 1 + 89119905 + 11989021199051199093 (119905) = minus2 minus 139119905 + 1198903119905(56)






1199091 (119905) = 13 + 19119905 + 1271199052 + 22431199053 + 17291199054 + 2109351199055+ 2984151199056 + 420667151199057 + 162001451199058+ 21674039151199059 + 119874 (11990510)






(60)




where


(62)


119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 + ℎ [1198961 + 21198962 + 21198963 + 1198964]6 (63)


Thus

119860 (119905) 119909119899+1 = [119860 (119905) minus (ℎ + ℎ22 + ℎ36 + ℎ424)119861 (119905)] 119909119899 (65)












(67)




119909119899 =

((


((


(69)

where 119902 isin C119899(119905)












Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of






(48)










(2 3 21 0 minus29 7 5 )(119890111989021198903) = (000) (55)


1199091 (119905) = 13 + 19119905 + 11989011199051199092 (119905) = 1 + 89119905 + 11989021199051199093 (119905) = minus2 minus 139119905 + 1198903119905(56)






1199091 (119905) = 13 + 19119905 + 1271199052 + 22431199053 + 17291199054 + 2109351199055+ 2984151199056 + 420667151199057 + 162001451199058+ 21674039151199059 + 119874 (11990510)






(60)




where


(62)


119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 + ℎ [1198961 + 21198962 + 21198963 + 1198964]6 (63)


Thus

119860 (119905) 119909119899+1 = [119860 (119905) minus (ℎ + ℎ22 + ℎ36 + ℎ424)119861 (119905)] 119909119899 (65)












(67)




119909119899 =

((


((


(69)

where 119902 isin C119899(119905)












Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of









(60)




where


(62)


119860 (119905) 119909119899+1 = 119860 (119905) 119909119899 + ℎ [1198961 + 21198962 + 21198963 + 1198964]6 (63)


Thus

119860 (119905) 119909119899+1 = [119860 (119905) minus (ℎ + ℎ22 + ℎ36 + ℎ424)119861 (119905)] 119909119899 (65)












(67)




119909119899 =

((


((


(69)

where 119902 isin C119899(119905)












Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of












(67)




119909119899 =

((


((


(69)

where 119902 isin C119899(119905)












Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of















Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of











Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of




numerical methods for a class of differential algebraic...

Documents