taylor polynomial solutions of general linear differential–difference equations with variable...

Applied Mathematics and Computation 174 (2006) 1526–1538

www.elsevier.com/locate/amc

Taylor polynomial solutions of generallinear differential–difference equations

with variable coefficients

Mehmet Sezer a, Ays�egul Akyuz-Das�cıoglu b,*

a Department of Mathematics, Faculty of Science, Mugla University, Mugla, Turkeyb Department of Mathematics, Faculty of Science, Pamukkale University, Denizli, Turkey

Abstract

In this paper, a Taylor method is developed to find an approximate solution of high-order linear differential–difference equations with variable coefficients under the mixedconditions. The solution is obtained in terms of Taylor polynomials. Examples are pre-sented which illustrate the pertinent features of the method.� 2005 Published by Elsevier Inc.

Keywords: Differential and difference equations; Taylor polynomials

1. Introduction

Taylor and Chebyshev matrix methods for the approximate solutions of dif-ferential equations have been presented in many papers [1–5]. In this paper,

0096-3003/$ - see front matter � 2005 Published by Elsevier Inc.doi:10.1016/j.amc.2005.07.002

* Corresponding author.E-mail address: [email protected] (A. Akyuz-Das�cıoglu).

mailto:[email protected]

M. Sezer, A. Akyuz-Das�cıoglu / Appl. Math. Comput. 174 (2006) 1526–1538 1527

these methods are developed and applied to high-order general linear differen-tial–difference equation with variable coefficients, which is given in [6, p. 229].Xm

k¼0

Xp

j¼0

PkjðxÞyðkÞðx� skjÞ ¼ f ðxÞ; skj P 0 ð1Þ

with the mixed conditionsXm�1

k¼0

XRr¼1

crikyðkÞðcrÞ ¼ ki; i ¼ 1; 2; . . . ;m; a 6 cr 6 b ð2Þ

and the solution is expressed as the Taylor polynomial

yðxÞ ¼XNn¼0

anðx� cÞn; a 6 x; c 6 b; ð3Þ

so that the Taylor coefficients to be determined are

an ¼yðnÞðcÞn!

; n ¼ 0; 1; . . . ;N . ð4Þ

Here Pkj(x) and f(x) are functions that have suitable derivatives on an intervala 6 x 6 b and crik, cr, c and skj are suitable coefficients.

2. Fundamental matrix relations

Let us convert the expressions defined in (1)–(3) to the matrix forms.Now, let us assume that the functions y(x) and its kth derivative with respect

to x, respectively, can be expanded to Taylor series about x = c in the forms

yðxÞ ¼X1n¼0

anðx� cÞn; an ¼yðnÞðcÞn!

ð5Þ

and

yðkÞðxÞ ¼X1n¼0

aðkÞn ðx� cÞn; ð6Þ

where for k = 0, y(0)(x) = y(x) and an ¼ að0Þn .First, let us derive the expression (6) with respect to x and then put

n ! n + 1:

yðkþ1ÞðxÞ ¼X1n¼1

naðkÞn ðx� cÞn�1 ¼X1n¼0

ðnþ 1ÞaðkÞnþ1ðx� cÞn. ð7Þ

From (6), it is clear that

yðkþ1ÞðxÞ ¼X1n¼0

aðkþ1Þn ðx� cÞn. ð8Þ

1528 M. Sezer, A. Akyuz-Das�cıoglu / Appl. Math. Comput. 174 (2006) 1526–1538

Using the relations (7) and (8), we have the recurrence relation between thecoefficients aðkÞn and aðkþ1Þ

n of y(k)(x) and y(k+1)(x)

aðkþ1Þn ¼ ðnþ 1ÞaðkÞnþ1; n; k ¼ 0; 1; 2; . . . ð9Þ

Now let us take n = 0,1, . . .,N and assume aðkÞn ¼ 0 for n > N. Then the system(9) can be transformed into the matrix form

Aðkþ1Þ ¼ MAðkÞ; k ¼ 0; 1; 2; . . . ; ð10Þwhere

AðkÞ ¼

aðkÞ0

aðkÞ1

..

.

aðkÞN

2666664

3777775; M ¼

0 1 0 � � � 0

0 0 2 � � � 0

..

. ... ..

. . .. ..

.

0 0 0 � � � N

0 0 0 � � � 0

26666664

37777775.

For k = 0,1,2, . . ., it follows from relation (10) that

AðkÞ ¼ MkA; ð11Þwhere clearly

Að0Þ ¼ A ¼ a0 a1 � � � aN½ �T.

Consequently, the matrix equation (11) gives a relation between the Taylorcoefficients matrix A of y(x) and the Taylor coefficients matrix A(k) of thekth derivative of y(x).

In addition, the solution expressed by (3) and its derivatives can be writtenin the matrix forms

yðxÞ ¼ XA and yðkÞðxÞ ¼ XAðkÞ ð12Þor using the relation in (11)

yðkÞðxÞ ¼ XMkA; ð13Þwhere

X ¼ 1 ðx� cÞ ðx� cÞ2 � � � ðx� cÞN� �

.

On the other hand, we can write the expression y(k)(x � skj) as

yðkÞðx� skjÞ ¼XNn¼0

aðkÞn x� c� skj� �n ¼ XN

n¼0

Xn

q¼0

n

q

� �x� cð Þn�q �skj

� �qaðkÞn

or in the matrix forms

yðkÞðx� skjÞ ¼ XT ðskjÞAðkÞ ð14Þ


in which T(skj) is a unit matrix for skj = 0, and for skj 5 0

T ðskjÞ ¼

0

0

� ��skj� �0 1

1

� ��skj� �1 2

2

� ��skj� �2 � � �

N

N

� ��skj� �N

01

0

� ��skj� �0 2

1

� ��skj� �1 � � �

N

N � 1

� ��skj� �N�1

0 02

0

� ��skj� �0 � � �

N

N � 2

� ��skj� �N�2

..

. ... ..

. . .. ..

.

0 0 0 � � �N

0

� ��skj� �0

266666666666666664

377777777777777775

.

In the similar way, the matrix representation of (x � c)iy(k)(x � skj) becomes

ðx� cÞiyðkÞðx� skjÞ ¼XNn¼0

Xn

q¼0

n

q

� �x� cð Þn�qþi �skj

� �qaðkÞn ¼ XIiT ðskjÞAðkÞ

or from (11)

ðx� cÞiyðkÞðx� skjÞ ¼ XIiT ðskjÞMkA; i ¼ 0; 1; . . . ;N ; ð15Þwhere

I0 ¼

1 0 � � � 0

0 1 � � � 0

..

. ... . .

. ...

0 0 � � � 1

266664

377775; I1 ¼

0 0 � � � 0 0

1 0 � � � 0 0

0 1 � � � 0 0

..

. ... . .

. ... ..

.

0 0 � � � 1 0

26666664

37777775;

I2 ¼

0 0 � � � 0 0 0

0 0 � � � 0 0 0

1 0 � � � 0 0 0

0 1 � � � 0 0 0

..

. ... . .

. ... ..

. ...

0 0 � � � 1 0 0

2666666664

3777777775; . . . ;

I i ¼

0 0 � � � 0 0 � � � 0

..

. ... ..

. ... ..

.

1 0 � � � 0 0 � � � 0

0 1 � � � 0 0 � � � 0

..

. ... . .

. ... ..

. ...

0 0 � � � 1 0 � � � 0

26666666664

37777777775; . . . ; IN ¼

0 0 � � � 0

0 0 � � � 0

..

. ... ..

.

1 0 � � � 0

266664

377775.


Besides, we assume that the function f(x) can be expanded as

f ðxÞ ¼XNn¼0

fnðx� cÞn; f n ¼f ðnÞðcÞn!

.

Then the matrix representation of f(x) becomes

f ðxÞ ¼ XF ; ð16Þwhere F ¼ ½ f0 f1 � � � fN �T.

3. Method of solution

We now ready to construct the fundamental matrix equation correspondingto Eq. (1). For this purpose, we first reduce Eq. (1) to the form

Xmk¼0

Xp

j¼0

XNi¼0

pikjðx� cÞiyðkÞðx� skjÞ ¼ f ðxÞ; ð17Þ

so that

PkjðxÞ ¼XNi¼0

pikj x� cð Þi; pikj ¼P ðiÞkj ðcÞi!

.

Substituting the matrix relations (15) and (16) into Eq. (17) and simplifying, weobtain the fundamental matrix equation

Xmk¼0

Xp

j¼0

XNi¼0

pikjI iT ðskjÞMkA ¼ F ; ð18Þ

which corresponds to a system of (N + 1) algebraic equations for the (N + 1)unknown coefficients a0,a1, . . .,aN . Briefly, we can write Eq. (18) in theform

WA ¼ F or ½W ; F �; ð19Þso that

W ¼ wnh½ � ¼Xmk¼0

Xp

j¼0

XNi¼0

pikjI iT ðskjÞMk; n; h ¼ 0; 1; . . . ;N .

We can obtain the corresponding matrix form for the mixed condition (2) as

Xm�1

k¼0

XRr¼0

crikCrMkA ¼ ki; i ¼ 1; 2; . . . ;m ð20Þ


by means of the relation (13) so that

Cr ¼ 1 ðcr � cÞ ðcr � cÞ2 . . . ðcr � cÞN� �

; r ¼ 0; 1; . . . ;R; a 6 cr 6 b.

Briefly, the matrix form for (2) is

UiA ¼ ki or ½Ui; ki�; i ¼ 1; 2; . . . ;m; ð21Þwhere

Ui ¼Xm�1

k¼0

XRr¼0

crikCrMk ¼ ½uij�; i ¼ 1; 2; . . . ;m; j ¼ 0; 1; . . . ;N

or clearly

Ui ¼ ½ ui0 ui1 . . . uiN �.

To obtain the solution of Eq. (1) under the mixed conditions (2), by replacingthe m rows matrices (21) by the last m rows of the matrix (19), we have the re-quired augmented matrix [1–5].

eW ; eFh i¼

w00 w01 . . . w0N ; f0w10 w11 . . . w1N ; f1

..

. ... ..

. ... ..

.

wN�m;0 wN�m;1 . . . wN�m;N ; fN�m

u10 u11 . . . u1N ; k1u20 u21 . . . u2N ; k2

..

. ... ..

. ... ..

.

um0 um1 . . . umN ; km

26666666666666664

37777777777777775

. ð22Þ

If rank eW ¼ rank½ eW ; eF � ¼ N þ 1, then we can write

A ¼ ð eW Þ�1eF . ð23ÞThus the coefficients an, n = 0,1, . . .,N are uniquely determined by Eq. (23).Also, by means of system (19) we may obtain the particular solutions.

4. Examples

Example 1. Let us illustrate the method by means of the differential–differenceequation

y 0ðxÞ � x2yðxÞ � xy 0ðx� 1Þ þ 2yðx� 2Þ ¼ �x4 � x3 þ x2 � 3xþ 3

with the condition y(�1) = �1 and approximate the solution y(x) by thepolynomial


yðxÞ ¼X2

n¼0

anxn;

where �2 6 x 6 0, m = 1, p = 1, N = 2, c = 0, P00(x) = �x2, s00 = 0;P01(x) = 2, s01 = 2; P10(x) = 1, s10 = 0; P11(x) = �x, s11 = 1.

We first reduce this equation, from Eq. (18), to the matrix form

X1

k¼0

X1

j¼0

X2

i¼0

pikjI iT ðskjÞMkA ¼ F .

The quantities in this equation are computed as

p000 ¼ 0; p100 ¼ 0; p200 ¼ �1; p010 ¼ 1; p110 ¼ 0; p210 ¼ 0;

p001 ¼ 2; p101 ¼ 0; p201 ¼ 0; p011 ¼ 0; p111 ¼ �1; p211 ¼ 0;

F ¼ ½3 � 3 1�T; A ¼ ½a0 a1 a2�T; T ðs00Þ ¼ T ðs10Þ ¼ T ð0Þ ¼ I0;

T ðs01Þ ¼ T ð2Þ ¼1 �2 4

0 1 �4

0 0 1

264

375; T ðs11Þ ¼ T ð1Þ ¼

1 �1 1

0 1 �2

0 0 1

264

375;

I0 ¼1 0 0

0 1 0

0 0 1

264

375; I1 ¼

0 0 0

1 0 0

0 1 0

264

375; I2 ¼

0 0 0

0 0 0

1 0 0

264

375; M ¼

0 1 0

0 0 2

0 0 0

264

375.

Hence the fundamental matrix equation becomes

�I2T ðs00Þ þ I0T ðs10ÞM þ 2I0T ðs01Þ � I1T ðs11ÞMf gA ¼ F

and the augmented matrix

W ; F½ � ¼2 �3 8 ; 3

0 1 �4 ; �3

�1 0 0 ; 1

264

375.

The augmented matrix based on the condition y(�1) = �1 is obtained as

eW ; eFh i¼

2 �3 8 ; 3

0 1 �4 ; �3

1 �1 1 ; �1

264

375.

Solving this system, the Taylor coefficients are obtained as

a0 ¼ �1; a1 ¼ 1; a2 ¼ 1


and thereby the polynomial solution becomes

yðxÞ ¼ �1þ xþ x2.

Note that from [W;F] we obtain the same result and taking N > 2 we can alsoobtain the same solution.

Example 2. Let us now solve the problem

y00ðxÞ � xy 0ðx� 1Þ þ yðx� 2Þ ¼ �x2 � 2xþ 5;

yð0Þ ¼ �1; y0ð�1Þ ¼ �2; �2 6 x 6 0.

We want to find the solution in the form

yðxÞ ¼X3

n¼0

anxn;

where c = 0, N = 3. The quantities in the given equation are

P 00ðxÞ ¼ 1; s00 ¼ 2; P 10ðxÞ ¼ �x; s10 ¼ 1; P 20ðxÞ ¼ 1; s20 ¼ 0.

The matrix form of equation is

X2

k¼0

X0

j¼0

X3

i¼0

pikjI iT ðskjÞMkA ¼ F ;

where

p000 ¼ 1; p100 ¼ 0; p200 ¼ 0; p300 ¼ 0;

p010 ¼ 0; p110 ¼ �1; p210 ¼ 0; p310 ¼ 0;

p020 ¼ 1; p120 ¼ 0; p220 ¼ 0; p320 ¼ 0

or simplifying

I0T ðs00Þ � I1T ðs10ÞM þ I0T ðs20ÞM2� �

A ¼ F ;

where

I0 ¼

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

26664

37775; I1 ¼

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

26664

37775; M ¼

0 1 0 0

0 0 2 0

0 0 0 3

0 0 0 0

26664

37775;

T ðs00Þ ¼ T ð2Þ ¼

1 �2 4 �8

0 1 �4 12

0 0 1 �6

0 0 0 1

26664

37775; T ðs10Þ ¼ T ð1Þ ¼

1 �1 1 �1

0 1 �2 3

0 0 1 �3

0 0 0 1

26664

37775;

T ðs20Þ ¼ T ð0Þ ¼ I0; F ¼ ½5 � 2 � 1 0�T.


Substituting these quantities in the previous equation, the augmented matrixfor WA = F is obtained as

W ; F½ � ¼

1 �2 6 �8 ; 5

0 0 �2 15 ; �2

0 0 �1 0 ; �1

0 0 0 �2 ; 0

266664

377775.

The augmented matrix for the given problem becomes

eW ; eFh i¼

1 �2 6 �8 ; 5

0 0 �2 15 ; �2

1 0 0 0 ; �1

0 1 �2 3 ; �2

266664

377775.

Solving this system, we have the polynomial solution

yðxÞ ¼ x2 � 1.

If F = 0, that is, the problem is homogenous, then we obtain

yðxÞ ¼ a1ðxþ 2Þ;from [W; 0], here a1 is the arbitrary constant.

Note from [W;F] that we can find a solution in the form

yðxÞ ¼ a1ðxþ 2Þ þ x2 � 1.

In addition, it is obtained the same results respectively both homogenous andnon-homogenous cases for N P 2.

Example 3. Let us consider the problem

y000ðxÞ � cosðxÞy0ðxÞ � sinðxÞy0 x� p2

þ

ffiffiffi2

py x� p

4

¼ sinðxÞ � 2 cosðxÞ � 1;

yð0Þ ¼ 0; y0ð0Þ ¼ 1; y00ð0Þ ¼ 0; � p26 x 6 0

and approximate the solution

yðxÞ ¼X5

n¼0

anxn.

The given equation is in the form

X3

k¼0

X1

j¼0

PkjðxÞyðkÞðx� skjÞ ¼ f ðxÞ;


where

P 00ðxÞ ¼ 0; s00 ¼ 0; P 01ðxÞ ¼ffiffiffi2

p; s01 ¼

p4;

P 10ðxÞ ¼ � cos x; s10 ¼ 0; P 11ðxÞ ¼ � sin x; s11 ¼p2;

P 20ðxÞ ¼ 0; s20 ¼ 0; P 21ðxÞ ¼ 0; s21 ¼ 0;

P 30ðxÞ ¼ 1; s30 ¼ 0; P 31ðxÞ ¼ 0; s31 ¼ 0.

The matrix form of this problem is

X3

k¼0

X1

j¼0

X5

i¼0

pikjI iT ðskjÞMkA ¼ F ;

where

pi00 ¼ pi20 ¼ pi21 ¼ pi31 ¼ 0 ði ¼ 0; 1; . . . ; 5Þ;

p001 ¼ffiffiffi2

p; p101 ¼ p201 ¼ p301 ¼ p401 ¼ p501 ¼ 0;

p010 ¼ �1; p110 ¼ 0; p210 ¼1

2; p310 ¼ 0; p410 ¼ � 1

24; p510 ¼ 0;

p011 ¼ 0; p111 ¼ �1; p211 ¼ 0; p311 ¼1

6; p411 ¼ 0; p511 ¼ � 1

120;

p030 ¼ 1; p130 ¼ p230 ¼ p330 ¼ p430 ¼ p530 ¼ 0;

T ðs00Þ ¼ T ðs10Þ ¼ T ðs20Þ ¼ T ðs30Þ ¼ T ðs21Þ ¼ T ðs31Þ ¼ I0;

Tp2

¼

1 � p2

p2

4� p3

8p4

16� p5

32

0 1 �p 3p2

4p3

25p4

16

0 0 1 � 3p2

3p2

2� 5p3

4

0 0 0 1 �2p 5p2

2

0 0 0 0 1 � 5p2

0 0 0 0 0 1

2666666666664

3777777777775; F ¼

�3

1

1

� 16

� 112

1120

266666666664

377777777775;

Tp4

¼

1 � p4

p2

16� p3

64p4

256� p5

1024

0 1 � p2

3p2

16� p3

165p4

256

0 0 1 � 3p4

3p2

8� 5p3

32

0 0 0 1 �p 5p2

8

0 0 0 0 1 � 5p4

0 0 0 0 0 1

266666666664

377777777775; M ¼

0 1 0 0 0 0

0 0 2 0 0 0

0 0 0 3 0 0

0 0 0 0 4 0

0 0 0 0 0 5

0 0 0 0 0 0

26666666664

37777777775;


I0 ¼

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

26666666664

37777777775; I1 ¼

0 0 0 0 0 0

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

26666666664

37777777775;

I2 ¼

0 0 0 0 0 0

0 0 0 0 0 0

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

26666666664

37777777775; I3 ¼

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

26666666664

37777777775;

I4 ¼

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

1 0 0 0 0 0

0 1 0 0 0 0

26666666664

37777777775; I5 ¼

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

1 0 0 0 0 0

26666666664

37777777775.

Substituting these values in equation and simplifying, we have the fundamentalmatrix equation:ffiffiffi

2p

Tp4

þ �I0 þ

1

2I2 �

1

24I4

� �M

�

þ �I1 þ1

6I3 �

1

120I5

� �T

p2

M þM3

A ¼ F .

After the ordinary operations, the augmented matrix [W;F] is obtained, andaugmented matrices for the conditions become

U 1; k1½ � ¼ ½1 0 0 0 0 0 ; 0�;

U 2; k2½ � ¼ ½0 1 0 0 0 0 ; 1�;

U 3; k3½ � ¼ ½0 0 2 0 0 0 ; 0�.Therefore the augmented matrix for the problem ½ eW ; eF � is gained. Solving thissystem, the approximate solution in terms of Taylor polynomial are obtainedas

yðxÞ ffi x� 0:166712x3 þ 0:000118x4 þ 0:007794x5.

Table 1Comparison of the absolute errors

xi N = 5 N = 8 N = 10 N = 12 N = 15

�p2 0.001523 0.000031 8.233735E�6 7.996475E�7 3.408115E�7�p3 0.000603 0.000014 4.927905E�6 7.107061E�6 6.002942E�6�p4 0.000192 4.510809E�6 1.67317E�6 2.792385E�6 2.39564E�6�p6 0.000034 7.929352E�7 3.092988E�7 5.73252E�7 4.969168E�70 0 0 0 0 0


In addition, for N = 8 we have the Taylor coefficients

A ¼ ½0 1 0 � 0:166668 2:885E� 6 0:008324 5:815E� 6

� 0:000195 � 6:56E� 6�

and we compare the absolute errors for different N in Table 1, since the exactsolution of this problem is sin(x).

5. Conclusions

The method presented in this study is useful in finding approximate and alsoexact solutions of certain differential–difference equations. In the cases m = 0and skj = 0, the method can be used respectively for difference equation anddifferential equation.

Differential–difference equations with variable coefficients are usually diffi-cult to solve analytically. In this case, the presented method is required forthe approximate solution.

On the other hand, it is observed that the method has the best advantagewhen the known functions in equation can be expanded to Taylor series withconverge rapidly (about x = c, �skj 6 c 6 0).

Also, the method can be used to solve the Fredholm integral and integro-differential–difference equations.

References

[1] M. Sezer, A method for the approximate solution of the second order linear differentialequations in terms of Taylor polynomials, Int. J. Math. Educ. Sci. Technol. 27 (6) (1996) 821–834.

[2] M. Sezer, M. Kaynak, Chebyshev polynomial solutions of linear differential equations, Int. J.Math. Educ. Sci. Technol. 27 (4) (1996) 607–618.

[3] S�. Nas, S. Yalcınbas�, M. Sezer, A Taylor polynomial approach for solving high-order linearFredholm integro-differential equations, Int. J. Math. Educ. Sci. Technol. 31 (2) (2000) 213–225.


[4] A. Akyuz, M. Sezer, A Chebyshev collocation method for the solution of linear integro-differential equations, Int. J. Comput. Math. 72 (4) (1999) 491–507.

[5] A. Akyuz, M. Sezer, Chebyshev polynomial solutions of systems of high-order linear differentialequations with variable coefficients, Appl. Math. Comput. 144 (2003) 237–247.

[6] T.L. Saaty, Modern Nonlinear Equations, Dover Publications, New York, 1981.

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