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Taylor polynomial solutions of general linear differential–difference equations with variable coefficients
Applied Mathematics and Computation 174 (2006) 1526–1538
www.elsevier.com/locate/amc
Taylor polynomial solutions of general linear differential–difference equations with variable coefficients Mehmet Sezer a, Aysßegu¨l Akyu¨z-Dasßcıog˘lu
b,*
a
b
Department of Mathematics, Faculty of Science, Mug˘la University, Mug˘la, Turkey 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 highorder linear differential–difference equations with variable coefficients under the mixed conditions. The solution is obtained in terms of Taylor polynomials. Examples are presented which illustrate the pertinent features of the method. Ó 2005 Published by Elsevier Inc. Keywords: Differential and difference equations; Taylor polynomials
1. Introduction Taylor and Chebyshev matrix methods for the approximate solutions of differential equations have been presented in many papers [1–5]. In this paper,
*
Corresponding author. E-mail address: [email protected] (A. Akyu¨z-Dasßcıog˘lu).
0096-3003/$ - see front matter Ó 2005 Published by Elsevier Inc. doi:10.1016/j.amc.2005.07.002
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1527
these methods are developed and applied to high-order general linear differential–difference equation with variable coefficients, which is given in [6, p. 229]. p m X X k¼0
P kj ðxÞy ðkÞ ðx skj Þ ¼ f ðxÞ;
with the mixed conditions m1 X R X crik y ðkÞ ðcr Þ ¼ ki ; k¼0
skj P 0
ð1Þ
j¼0
i ¼ 1; 2; . . . ; m; a 6 cr 6 b
ð2Þ
r¼1
and the solution is expressed as the Taylor polynomial N X n an ðx cÞ ; a 6 x; c 6 b; yðxÞ ¼
ð3Þ
n¼0
so that the Taylor coefficients to be determined are an ¼
y ðnÞ ðcÞ ; n!
n ¼ 0; 1; . . . ; N .
ð4Þ
Here Pkj(x) and f(x) are functions that have suitable derivatives on an interval a 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 1 X y ðnÞ ðcÞ ð5Þ yðxÞ ¼ an ðx cÞn ; an ¼ n! n¼0 and y ðkÞ ðxÞ ¼
1 X
n
aðkÞ n ðx cÞ ;
ð6Þ
n¼0
where for k = 0, y(0)(x) = y(x) and an ¼ að0Þ n . First, let us derive the expression (6) with respect to x and then put n ! n + 1: 1 1 X X ðkÞ n1 n y ðkþ1Þ ðxÞ ¼ naðkÞ ¼ ðn þ 1Þanþ1 ðx cÞ . ð7Þ n ðx cÞ n¼1
From (6), it is clear that 1 X y ðkþ1Þ ðxÞ ¼ anðkþ1Þ ðx cÞn . n¼0
n¼0
ð8Þ
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1528
Using the relations (7) and (8), we have the recurrence relation between the ðkþ1Þ coefficients aðkÞ of y(k)(x) and y(k+1)(x) n and an ðkÞ
anðkþ1Þ ¼ ðn þ 1Þanþ1 ;
n; k ¼ 0; 1; 2; . . .
ð9Þ
aðkÞ n
¼ 0 for n > N. Then the system Now let us take n = 0, 1, . . ., N and assume (9) can be transformed into the matrix form Aðkþ1Þ ¼ MAðkÞ ;
k ¼ 0; 1; 2; . . . ;
ð10Þ
where 2 AðkÞ
ðkÞ
a0
2
3
6 ðkÞ 7 6a 7 6 1 7 ¼ 6 . 7; 6 . 7 4 . 5 ðkÞ
aN
3
0
1
0
60 6 6. . M ¼6 6. 6 40
0 .. .
2 .. .
0 0
0 0
7 7 7 7. 7 7 N 5 0
0
0
.. .
0 .. .
For k = 0, 1, 2, . . ., it follows from relation (10) that AðkÞ ¼ M k A;
ð11Þ
where clearly Að0Þ ¼ A ¼ ½ a0
a1
T
aN .
Consequently, the matrix equation (11) gives a relation between the Taylor coefficients matrix A of y(x) and the Taylor coefficients matrix A(k) of the kth derivative of y(x). In addition, the solution expressed by (3) and its derivatives can be written in the matrix forms yðxÞ ¼ XA
and
y ðkÞ ðxÞ ¼ XAðkÞ
ð12Þ
or using the relation in (11) y ðkÞ ðxÞ ¼ XM k A; where
X ¼ 1 ðx cÞ ðx cÞ2
ð13Þ
ðx cÞ
N
.
On the other hand, we can write the expression y(k)(x skj) as N N X n X n X q n nq ðkÞ ðkÞ y ðx skj Þ ¼ an x c skj ¼ skj aðkÞ ðx cÞ n q n¼0 n¼0 q¼0 or in the matrix forms y ðkÞ ðx skj Þ ¼ XT ðskj ÞAðkÞ
ð14Þ
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
in which T(skj) is a unit matrix for skj = 0, and for skj 5 0 2 0 0 6 0 skj 6 6 6 6 0 6 6 6 T ðskj Þ ¼ 6 6 0 6 6 6 .. 6 6 . 6 4 0
1
.. .
2 2 skj 2 1 2 skj 1 0 2 skj 0 .. .
0
0
1 1 0
skj skj
1 0
0
N
skj
1529
N
3
7 7 7 N1 7 7 skj 7 N 1 7 7 N N2 7. 7 skj 7 N 2 7 7 .. .. 7 7 . . 7 0 5 N skj 0
N N
In the similar way, the matrix representation of (x c)iy(k)(x skj) becomes N X n X q n ðkÞ ðx cÞi y ðkÞ ðx skj Þ ¼ ðx cÞnqþi skj aðkÞ n ¼ XI i T ðskj ÞA q n¼0 q¼0 or from (11) i
ðx cÞ y ðkÞ ðx skj Þ ¼ XI i T ðskj ÞM k A; where 2
1 60 6 I0 ¼ 6 6 .. 4. 0 2
3 0 07 7 .. 7 7; .5
0 1 .. . . . . 0
1
0 60 6 61 6 I2 ¼ 6 60 6 6 .. 4.
0 0 0 1 .. .
.. .
0 0 0 0 .. .
0 0 0 0 .. .
0 0 6 .. 6. 6 61 6 Ii ¼ 6 60 6 6 .. 4. 0
0 0 .. .
1 0 .. .
0 0 .. .
0 1 .. . . . . 0
0 0 0 0 .. .. . . 1 0
2
2
i ¼ 0; 1; . . . ; N ;
0 0 61 0 6 6 0 1 I1 ¼ 6 6. . 6. . 4. . 0 0 3 0 07 7 07 7 7; . . . ; 07 7 .. 7 .5 0 3 0 .. 7 .7 7 07 7 7; . . . ; 07 7 .. 7 .5
0
.. .
3 0 0 0 07 7 7 0 0 7; .. .. 7 7 . .5
1 0
2
ð15Þ
0 60 6 IN ¼ 6 6 .. 4.
0 0 .. .
3 0 07 7 .. 7 7. .5
1
0
0
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1530
Besides, we assume that the function f(x) can be expanded as f ðxÞ ¼
N X
n
fn ðx cÞ ;
f ðnÞ ðcÞ . n!
fn ¼
n¼0
Then the matrix representation of f(x) becomes f ðxÞ ¼ XF ; where F ¼ ½ f0
ð16Þ f1
T
fN .
3. Method of solution We now ready to construct the fundamental matrix equation corresponding to Eq. (1). For this purpose, we first reduce Eq. (1) to the form p X m X N X k¼0
i
pikj ðx cÞ y ðkÞ ðx skj Þ ¼ f ðxÞ;
ð17Þ
i¼0
j¼0
so that P kj ðxÞ ¼
N X
ðiÞ
pikj ðx
i
cÞ ;
i¼0
pikj
P kj ðcÞ ¼ . i!
Substituting the matrix relations (15) and (16) into Eq. (17) and simplifying, we obtain the fundamental matrix equation p X m X N X k¼0
j¼0
pikj I i T ðskj ÞM k A ¼ F ;
ð18Þ
i¼0
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 the form WA ¼ F or ½W ; F ;
ð19Þ
so that W ¼ ½wnh ¼
p X m X N X k¼0
j¼0
pikj I i T ðskj ÞM k ;
n; h ¼ 0; 1; . . . ; N .
i¼0
We can obtain the corresponding matrix form for the mixed condition (2) as m1 X R X k¼0
r¼0
crik C r M k A ¼ ki ;
i ¼ 1; 2; . . . ; m
ð20Þ
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1531
by means of the relation (13) so that C r ¼ 1 ðcr cÞ ðcr cÞ2
...
ðcr cÞ
N
;
r ¼ 0; 1; . . . ; R;
a 6 cr 6 b.
Briefly, the matrix form for (2) is U i A ¼ ki
½U i ; ki ;
or
i ¼ 1; 2; . . . ; m;
ð21Þ
where Ui ¼
m1 X R X k¼0
crik C r M k ¼ ½uij ;
i ¼ 1; 2; . . . ; m;
j ¼ 0; 1; . . . ; N
r¼0
or clearly U i ¼ ½ ui0
ui1
...
uiN .
To obtain the solution of Eq. (1) under the mixed conditions (2), by replacing the m rows matrices (21) by the last m rows of the matrix (19), we have the required augmented matrix [1–5]. 3 2 w00 w01 ... w0N ; f0 6 w w11 ... w1N ; f1 7 7 6 10 6 . .. 7 .. .. .. 7 6 . 6 . . 7 . . . 7 6 7 h i 6w 6 N m;0 wN m;1 . . . wN m;N ; fN m 7 e e ð22Þ W;F ¼6 7. 6 u10 u11 ... u1N ; k1 7 7 6 6 u u21 ... u2N ; k2 7 7 6 20 6 . .. 7 .. .. .. 7 6 . 4 . . 5 . . . um0 um1 ... umN ; km e ¼ rank½ W e ; Fe ¼ N þ 1, then we can write If rank W e Þ1 Fe . A ¼ ðW
ð23Þ
Thus 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–difference equation y 0 ðxÞ x2 yðxÞ xy 0 ðx 1Þ þ 2yðx 2Þ ¼ x4 x3 þ x2 3x þ 3 with the condition y(1) = 1 and approximate the solution y(x) by the polynomial
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yðxÞ ¼
2 X
an xn ;
n¼0
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 1 X 1 X 2 X k¼0
j¼0
pikj I i T ðskj ÞM k A ¼ F .
i¼0
The quantities in this equation are computed as p000 ¼ 0;
p100 ¼ 0;
p200 ¼ 1;
p001 ¼ 2;
p101 ¼ 0;
p201 ¼ 0;
p011 ¼ 0;
T
A ¼ ½a0
a1
F ¼ ½3
3
1 ; 2
1 2
6 T ðs01 Þ ¼ T ð2Þ ¼ 4 0 0 2
1
6 I0 ¼ 4 0 0
0 0
3
T
a2 ;
0 0
6 I1 ¼ 4 1 0 0 1
p110 ¼ 0;
p210 ¼ 0;
p111 ¼ 1;
p211 ¼ 0;
T ðs00 Þ ¼ T ðs10 Þ ¼ T ð0Þ ¼ I 0 ;
3
2
7 4 5; 1
1 0 2
7 1 0 5; 0 1
4
p010 ¼ 1;
0
3
7 0 5; 0
1 1
6 T ðs11 Þ ¼ T ð1Þ ¼ 4 0 0 2
0
6 I2 ¼ 4 0 1
0 0
3
7 0 0 5; 0 0
1 0 2
0
6 M ¼ 40 0
1
7 2 5; 1 1 0
fI 2 T ðs00 Þ þ I 0 T ðs10 ÞM þ 2I 0 T ðs01 Þ I 1 T ðs11 ÞM gA ¼ F
1
0
0
; ; ;
3 3 7 3 5. 1
The augmented matrix based on the condition y(1) = 1 is obtained as 2 3 2 3 8 ; 3 h i 7 e ; Fe ¼ 6 W 4 0 1 4 ; 3 5. 1 1 1 ; 1 Solving this system, the Taylor coefficients are obtained as a0 ¼ 1;
a1 ¼ 1;
a2 ¼ 1
3
7 0 2 5. 0 0
Hence the fundamental matrix equation becomes and the augmented matrix 2 2 3 8 6 ½W ; F ¼ 4 0 1 4
3
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1533
and thereby the polynomial solution becomes yðxÞ ¼ 1 þ x þ x2 . Note that from [W; F] we obtain the same result and taking N > 2 we can also obtain the same solution. Example 2. Let us now solve the problem y 00 ðxÞ xy 0 ðx 1Þ þ yðx 2Þ ¼ x2 2x þ 5; yð0Þ ¼ 1; y 0 ð1Þ ¼ 2; 2 6 x 6 0. We want to find the solution in the form yðxÞ ¼
3 X
an xn ;
n¼0
where c = 0, N = 3. The quantities in the given equation are P 00 ðxÞ ¼ 1; s00 ¼ 2;
P 10 ðxÞ ¼ x; s10 ¼ 1;
P 20 ðxÞ ¼ 1; s20 ¼ 0.
The matrix form of equation is 2 X 0 X 3 X k¼0
j¼0
pikj I i T ðskj ÞM k A ¼ F ;
i¼0
where p000 ¼ 1;
p100 ¼ 0;
p200 ¼ 0;
p010 ¼ 0;
p110 ¼ 1;
p020 ¼ 1;
p120 ¼ 0;
p300 ¼ 0;
p210 ¼ 0; p220 ¼ 0;
p310 ¼ 0; p320 ¼ 0
or simplifying I 0 T ðs00 Þ I 1 T ðs10 ÞM þ I 0 T ðs20 ÞM 2 A ¼ F ; where 3 0 07 7 7; 05 1 2 1 60 6 T ðs00 Þ ¼ T ð2Þ ¼ 6 40 0 2
1 60 6 I0 ¼ 6 40 0
0 1 0 0
0 0 1 0
T ðs20 Þ ¼ T ð0Þ ¼ I 0 ;
2
0 0 61 0 6 I1 ¼ 6 40 1 0 0 3 2 4 8 1 4 12 7 7 7; 0 1 6 5 0 0 1 F ¼ ½5
2
0 0 0 1
3 0 07 7 7; 05 0
2
1 0 0 0 1 60 6 T ðs10 Þ ¼ T ð1Þ ¼ 6 40 0
1
0T .
0 60 6 M ¼6 40 0 2
0 2 0 0 1 1 0 0
3 0 07 7 7; 35 0 3 1 1 2 3 7 7 7; 1 3 5 0 1
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Substituting these quantities in the for WA = F is obtained as 2 1 2 6 8 6 6 0 0 2 15 ½W ; F ¼ 6 6 0 0 1 0 4 0
0
0
2
previous equation, the augmented matrix 5
;
3
7 ; 2 7 7. ; 1 7 5 0
;
The augmented matrix for the given problem becomes 2 3 1 2 6 8 ; 5 7 h i 6 6 0 0 2 15 ; 2 7 7. e ; Fe ¼ 6 W 61 0 0 0 ; 1 7 4 5 0
1
2
3
; 2
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 a1 ðx þ 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 a1 ðx þ 2Þ þ x2 1. In addition, it is obtained the same results respectively both homogenous and non-homogenous cases for N P 2. Example 3. Let us consider the problem p pffiffiffi p
y 000 ðxÞ cosðxÞy 0 ðxÞ sinðxÞy 0 x þ 2y x ¼ sinðxÞ 2 cosðxÞ 1; 2 4 p y 00 ð0Þ ¼ 0; 6x60 yð0Þ ¼ 0; y 0 ð0Þ ¼ 1; 2 and approximate the solution yðxÞ ¼
5 X
an xn .
n¼0
The given equation is in the form 3 X 1 X k¼0
j¼0
P kj ðxÞy ðkÞ ðx skj Þ ¼ f ðxÞ;
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1535
where P 00 ðxÞ ¼ 0; s00 ¼ 0;
P 01 ðxÞ ¼
pffiffiffi p 2; s01 ¼ ; 4
P 20 ðxÞ ¼ 0; s20 ¼ 0;
p P 11 ðxÞ ¼ sin x; s11 ¼ ; 2 P 21 ðxÞ ¼ 0; s21 ¼ 0;
P 30 ðxÞ ¼ 1; s30 ¼ 0;
P 31 ðxÞ ¼ 0; s31 ¼ 0.
P 10 ðxÞ ¼ cos x; s10 ¼ 0;
The matrix form of this problem is 3 X 1 X 5 X k¼0
j¼0
pikj I i T ðskj ÞM k A ¼ F ;
i¼0
where pi00 ¼ pi20 ¼ pi21 ¼ pi31 ¼ 0 ði ¼ 0; 1; . . . ; 5Þ; pffiffiffi p001 ¼ 2; p101 ¼ p201 ¼ p301 ¼ p401 ¼ p501 ¼ 0; p110 ¼ 0;
1 p210 ¼ ; 2
p011 ¼ 0;
p111 ¼ 1;
p211 ¼ 0;
p030 ¼ 1;
p130 ¼ p230 ¼ p330 ¼ p430
p010 ¼ 1;
p310 ¼ 0;
p410 ¼
1 ; 24
1 p311 ¼ ; p411 ¼ 0; 6 5 ¼ p30 ¼ 0;
p510 ¼ 0;
p511 ¼
1 ; 120
T ðs00 Þ ¼ T ðs10 Þ ¼ T ðs20 Þ ¼ T ðs30 Þ ¼ T ðs21 Þ ¼ T ðs31 Þ ¼ I 0 ; 2
6 60 6 6 p 6 0 6 T ¼6 6 2 60 6 6 40 0 2
p2 4
p8
p4 16
1
p
3p2 4
p3 2
0
1
3p 2
3p2 2
0
0
1
2p
0
0
0
1
0
0
0
0
1 p2
1
6 60 6 p 6 60 T ¼6 6 4 60 6 6 40 0
3
3
5
p32
7 7 7 7 5p3 7 4 7 7; 5p2 7 7 2 7 7 5p 2 5 5p4 16
2
p2 16
p64
p4 256
5
1
p2
3p2 16
p16
5p4 256
0
1
3p 4
3p2 8
5p 32
p 1024
3
0
0
1
p
0
0
0
1
5p2 8 5p4
0
0
0
0
1
3 7 7 7 7 7 7; 7 7 7 7 5
3
3
7 6 6 1 7 7 6 7 6 6 1 7 7 F ¼6 6 1 7; 6 67 7 6 6 1 7 4 12 5 1 120
1
p4
3
3
2
0 1 0 0
6 60 6 60 6 M ¼6 60 6 6 40
0 2 0 0 0 3 0 0 0 0 0 0
0 0 0 0
0 0
3
7 0 07 7 0 07 7 7; 4 07 7 7 0 55 0 0
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2
1
6 60 6 60 6 I0 ¼ 6 60 6 60 4 0 2 0 6 60 6 61 6 I2 ¼ 6 60 6 60 4 0 2 0 6 60 6 60 6 I4 ¼ 6 60 6 61 4 0
0 0
0
0
1 0
0
0
0 1
0
0
0 0
1
0
0 0
0
1
0 0
0
0
0 0
0
0
0 0
0
0
0 0
0
0
1 0
0
0
0 1
0
0
0 0
1
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
3
2
7 07 7 07 7 7; 07 7 07 5 1 3 0 7 07 7 07 7 7; 07 7 07 5 0 3 0 7 07 7 07 7 7; 07 7 07 5 0
0
6 61 6 60 6 I1 ¼ 6 60 6 60 4 0 2 0 6 60 6 60 6 I3 ¼ 6 61 6 60 4 0 2 0 6 60 6 60 6 I5 ¼ 6 60 6 60 4 1
0
0
0 0
0
0
0 0
1
0
0 0
0
1
0 0
0
0
1 0
0
0
0 1
0
0
0 0
0
0
0 0
0
0
0 0
0
0
0 0
1
0
0 0
0
1
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
3
7 07 7 07 7 7; 07 7 07 5 0 3 0 7 07 7 07 7 7; 07 7 07 5 0 3 0 7 07 7 07 7 7. 07 7 07 5 0
Substituting these values in equation and simplifying, we have the fundamental matrix equation: pffiffiffi p
1 1 2T þ I 0 þ I 2 I 4 M 4 2 24
1 1 p 3 I5 T þ I 1 þ I 3 M þ M A ¼ F. 6 120 2 After the ordinary operations, the augmented matrix [W; F] is obtained, and augmented matrices for the conditions become ½U 1 ; k1 ¼ ½1
0
0
0
0 0
;
0;
½U 2 ; k2 ¼ ½0
1
0
0
0 0
;
1;
½U 3 ; k3 ¼ ½0
0
2
0
0 0
;
0.
e ; Fe is gained. Solving this Therefore the augmented matrix for the problem ½ W system, the approximate solution in terms of Taylor polynomial are obtained as yðxÞ ffi x 0:166712x3 þ 0:000118x4 þ 0:007794x5 .
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1537
Table 1 Comparison of the absolute errors xi
N=5
N=8
N = 10
N = 12
N = 15
p 2 p 3 p 4 p 6
0.001523 0.000603 0.000192 0.000034 0
0.000031 0.000014 4.510809E6 7.929352E7 0
8.233735E6 4.927905E6 1.67317E6 3.092988E7 0
7.996475E7 7.107061E6 2.792385E6 5.73252E7 0
3.408115E7 6.002942E6 2.39564E6 4.969168E7 0
0
In addition, for N = 8 we have the Taylor coefficients A ¼ ½0 1
0
0:000195
0:166668
2:885E 6 0:008324
5:815E 6
6:56E 6
and we compare the absolute errors for different N in Table 1, since the exact solution of this problem is sin(x).
5. Conclusions The method presented in this study is useful in finding approximate and also exact solutions of certain differential–difference equations. In the cases m = 0 and skj = 0, the method can be used respectively for difference equation and differential equation. Differential–difference equations with variable coefficients are usually difficult to solve analytically. In this case, the presented method is required for the approximate solution. On the other hand, it is observed that the method has the best advantage when the known functions in equation can be expanded to Taylor series with converge rapidly (about x = c, skj 6 c 6 0). Also, the method can be used to solve the Fredholm integral and integrodifferential–difference equations. References [1] M. Sezer, A method for the approximate solution of the second order linear differential equations 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 ß . Nas, S. Yalc¸ınbasß, M. Sezer, A Taylor polynomial approach for solving high-order linear Fredholm integro-differential equations, Int. J. Math. Educ. Sci. Technol. 31 (2) (2000) 213– 225.
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[4] A. Akyu¨z, M. Sezer, A Chebyshev collocation method for the solution of linear integrodifferential equations, Int. J. Comput. Math. 72 (4) (1999) 491–507. [5] A. Akyu¨z, M. Sezer, Chebyshev polynomial solutions of systems of high-order linear differential equations with variable coefficients, Appl. Math. Comput. 144 (2003) 237–247. [6] T.L. Saaty, Modern Nonlinear Equations, Dover Publications, New York, 1981.
www.elsevier.com/locate/amc
Taylor polynomial solutions of general linear differential–difference equations with variable coefficients Mehmet Sezer a, Aysßegu¨l Akyu¨z-Dasßcıog˘lu
b,*
a
b
Department of Mathematics, Faculty of Science, Mug˘la University, Mug˘la, Turkey 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 highorder linear differential–difference equations with variable coefficients under the mixed conditions. The solution is obtained in terms of Taylor polynomials. Examples are presented which illustrate the pertinent features of the method. Ó 2005 Published by Elsevier Inc. Keywords: Differential and difference equations; Taylor polynomials
1. Introduction Taylor and Chebyshev matrix methods for the approximate solutions of differential equations have been presented in many papers [1–5]. In this paper,
*
Corresponding author. E-mail address: [email protected] (A. Akyu¨z-Dasßcıog˘lu).
0096-3003/$ - see front matter Ó 2005 Published by Elsevier Inc. doi:10.1016/j.amc.2005.07.002
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1527
these methods are developed and applied to high-order general linear differential–difference equation with variable coefficients, which is given in [6, p. 229]. p m X X k¼0
P kj ðxÞy ðkÞ ðx skj Þ ¼ f ðxÞ;
with the mixed conditions m1 X R X crik y ðkÞ ðcr Þ ¼ ki ; k¼0
skj P 0
ð1Þ
j¼0
i ¼ 1; 2; . . . ; m; a 6 cr 6 b
ð2Þ
r¼1
and the solution is expressed as the Taylor polynomial N X n an ðx cÞ ; a 6 x; c 6 b; yðxÞ ¼
ð3Þ
n¼0
so that the Taylor coefficients to be determined are an ¼
y ðnÞ ðcÞ ; n!
n ¼ 0; 1; . . . ; N .
ð4Þ
Here Pkj(x) and f(x) are functions that have suitable derivatives on an interval a 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 1 X y ðnÞ ðcÞ ð5Þ yðxÞ ¼ an ðx cÞn ; an ¼ n! n¼0 and y ðkÞ ðxÞ ¼
1 X
n
aðkÞ n ðx cÞ ;
ð6Þ
n¼0
where for k = 0, y(0)(x) = y(x) and an ¼ að0Þ n . First, let us derive the expression (6) with respect to x and then put n ! n + 1: 1 1 X X ðkÞ n1 n y ðkþ1Þ ðxÞ ¼ naðkÞ ¼ ðn þ 1Þanþ1 ðx cÞ . ð7Þ n ðx cÞ n¼1
From (6), it is clear that 1 X y ðkþ1Þ ðxÞ ¼ anðkþ1Þ ðx cÞn . n¼0
n¼0
ð8Þ
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Using the relations (7) and (8), we have the recurrence relation between the ðkþ1Þ coefficients aðkÞ of y(k)(x) and y(k+1)(x) n and an ðkÞ
anðkþ1Þ ¼ ðn þ 1Þanþ1 ;
n; k ¼ 0; 1; 2; . . .
ð9Þ
aðkÞ n
¼ 0 for n > N. Then the system Now let us take n = 0, 1, . . ., N and assume (9) can be transformed into the matrix form Aðkþ1Þ ¼ MAðkÞ ;
k ¼ 0; 1; 2; . . . ;
ð10Þ
where 2 AðkÞ
ðkÞ
a0
2
3
6 ðkÞ 7 6a 7 6 1 7 ¼ 6 . 7; 6 . 7 4 . 5 ðkÞ
aN
3
0
1
0
60 6 6. . M ¼6 6. 6 40
0 .. .
2 .. .
0 0
0 0
7 7 7 7. 7 7 N 5 0
0
0
.. .
0 .. .
For k = 0, 1, 2, . . ., it follows from relation (10) that AðkÞ ¼ M k A;
ð11Þ
where clearly Að0Þ ¼ A ¼ ½ a0
a1
T
aN .
Consequently, the matrix equation (11) gives a relation between the Taylor coefficients matrix A of y(x) and the Taylor coefficients matrix A(k) of the kth derivative of y(x). In addition, the solution expressed by (3) and its derivatives can be written in the matrix forms yðxÞ ¼ XA
and
y ðkÞ ðxÞ ¼ XAðkÞ
ð12Þ
or using the relation in (11) y ðkÞ ðxÞ ¼ XM k A; where
X ¼ 1 ðx cÞ ðx cÞ2
ð13Þ
ðx cÞ
N
.
On the other hand, we can write the expression y(k)(x skj) as N N X n X n X q n nq ðkÞ ðkÞ y ðx skj Þ ¼ an x c skj ¼ skj aðkÞ ðx cÞ n q n¼0 n¼0 q¼0 or in the matrix forms y ðkÞ ðx skj Þ ¼ XT ðskj ÞAðkÞ
ð14Þ
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
in which T(skj) is a unit matrix for skj = 0, and for skj 5 0 2 0 0 6 0 skj 6 6 6 6 0 6 6 6 T ðskj Þ ¼ 6 6 0 6 6 6 .. 6 6 . 6 4 0
1
.. .
2 2 skj 2 1 2 skj 1 0 2 skj 0 .. .
0
0
1 1 0
skj skj
1 0
0
N
skj
1529
N
3
7 7 7 N1 7 7 skj 7 N 1 7 7 N N2 7. 7 skj 7 N 2 7 7 .. .. 7 7 . . 7 0 5 N skj 0
N N
In the similar way, the matrix representation of (x c)iy(k)(x skj) becomes N X n X q n ðkÞ ðx cÞi y ðkÞ ðx skj Þ ¼ ðx cÞnqþi skj aðkÞ n ¼ XI i T ðskj ÞA q n¼0 q¼0 or from (11) i
ðx cÞ y ðkÞ ðx skj Þ ¼ XI i T ðskj ÞM k A; where 2
1 60 6 I0 ¼ 6 6 .. 4. 0 2
3 0 07 7 .. 7 7; .5
0 1 .. . . . . 0
1
0 60 6 61 6 I2 ¼ 6 60 6 6 .. 4.
0 0 0 1 .. .
.. .
0 0 0 0 .. .
0 0 0 0 .. .
0 0 6 .. 6. 6 61 6 Ii ¼ 6 60 6 6 .. 4. 0
0 0 .. .
1 0 .. .
0 0 .. .
0 1 .. . . . . 0
0 0 0 0 .. .. . . 1 0
2
2
i ¼ 0; 1; . . . ; N ;
0 0 61 0 6 6 0 1 I1 ¼ 6 6. . 6. . 4. . 0 0 3 0 07 7 07 7 7; . . . ; 07 7 .. 7 .5 0 3 0 .. 7 .7 7 07 7 7; . . . ; 07 7 .. 7 .5
0
.. .
3 0 0 0 07 7 7 0 0 7; .. .. 7 7 . .5
1 0
2
ð15Þ
0 60 6 IN ¼ 6 6 .. 4.
0 0 .. .
3 0 07 7 .. 7 7. .5
1
0
0
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Besides, we assume that the function f(x) can be expanded as f ðxÞ ¼
N X
n
fn ðx cÞ ;
f ðnÞ ðcÞ . n!
fn ¼
n¼0
Then the matrix representation of f(x) becomes f ðxÞ ¼ XF ; where F ¼ ½ f0
ð16Þ f1
T
fN .
3. Method of solution We now ready to construct the fundamental matrix equation corresponding to Eq. (1). For this purpose, we first reduce Eq. (1) to the form p X m X N X k¼0
i
pikj ðx cÞ y ðkÞ ðx skj Þ ¼ f ðxÞ;
ð17Þ
i¼0
j¼0
so that P kj ðxÞ ¼
N X
ðiÞ
pikj ðx
i
cÞ ;
i¼0
pikj
P kj ðcÞ ¼ . i!
Substituting the matrix relations (15) and (16) into Eq. (17) and simplifying, we obtain the fundamental matrix equation p X m X N X k¼0
j¼0
pikj I i T ðskj ÞM k A ¼ F ;
ð18Þ
i¼0
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 the form WA ¼ F or ½W ; F ;
ð19Þ
so that W ¼ ½wnh ¼
p X m X N X k¼0
j¼0
pikj I i T ðskj ÞM k ;
n; h ¼ 0; 1; . . . ; N .
i¼0
We can obtain the corresponding matrix form for the mixed condition (2) as m1 X R X k¼0
r¼0
crik C r M k A ¼ ki ;
i ¼ 1; 2; . . . ; m
ð20Þ
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by means of the relation (13) so that C r ¼ 1 ðcr cÞ ðcr cÞ2
...
ðcr cÞ
N
;
r ¼ 0; 1; . . . ; R;
a 6 cr 6 b.
Briefly, the matrix form for (2) is U i A ¼ ki
½U i ; ki ;
or
i ¼ 1; 2; . . . ; m;
ð21Þ
where Ui ¼
m1 X R X k¼0
crik C r M k ¼ ½uij ;
i ¼ 1; 2; . . . ; m;
j ¼ 0; 1; . . . ; N
r¼0
or clearly U i ¼ ½ ui0
ui1
...
uiN .
To obtain the solution of Eq. (1) under the mixed conditions (2), by replacing the m rows matrices (21) by the last m rows of the matrix (19), we have the required augmented matrix [1–5]. 3 2 w00 w01 ... w0N ; f0 6 w w11 ... w1N ; f1 7 7 6 10 6 . .. 7 .. .. .. 7 6 . 6 . . 7 . . . 7 6 7 h i 6w 6 N m;0 wN m;1 . . . wN m;N ; fN m 7 e e ð22Þ W;F ¼6 7. 6 u10 u11 ... u1N ; k1 7 7 6 6 u u21 ... u2N ; k2 7 7 6 20 6 . .. 7 .. .. .. 7 6 . 4 . . 5 . . . um0 um1 ... umN ; km e ¼ rank½ W e ; Fe ¼ N þ 1, then we can write If rank W e Þ1 Fe . A ¼ ðW
ð23Þ
Thus 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–difference equation y 0 ðxÞ x2 yðxÞ xy 0 ðx 1Þ þ 2yðx 2Þ ¼ x4 x3 þ x2 3x þ 3 with the condition y(1) = 1 and approximate the solution y(x) by the polynomial
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yðxÞ ¼
2 X
an xn ;
n¼0
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 1 X 1 X 2 X k¼0
j¼0
pikj I i T ðskj ÞM k A ¼ F .
i¼0
The quantities in this equation are computed as p000 ¼ 0;
p100 ¼ 0;
p200 ¼ 1;
p001 ¼ 2;
p101 ¼ 0;
p201 ¼ 0;
p011 ¼ 0;
T
A ¼ ½a0
a1
F ¼ ½3
3
1 ; 2
1 2
6 T ðs01 Þ ¼ T ð2Þ ¼ 4 0 0 2
1
6 I0 ¼ 4 0 0
0 0
3
T
a2 ;
0 0
6 I1 ¼ 4 1 0 0 1
p110 ¼ 0;
p210 ¼ 0;
p111 ¼ 1;
p211 ¼ 0;
T ðs00 Þ ¼ T ðs10 Þ ¼ T ð0Þ ¼ I 0 ;
3
2
7 4 5; 1
1 0 2
7 1 0 5; 0 1
4
p010 ¼ 1;
0
3
7 0 5; 0
1 1
6 T ðs11 Þ ¼ T ð1Þ ¼ 4 0 0 2
0
6 I2 ¼ 4 0 1
0 0
3
7 0 0 5; 0 0
1 0 2
0
6 M ¼ 40 0
1
7 2 5; 1 1 0
fI 2 T ðs00 Þ þ I 0 T ðs10 ÞM þ 2I 0 T ðs01 Þ I 1 T ðs11 ÞM gA ¼ F
1
0
0
; ; ;
3 3 7 3 5. 1
The augmented matrix based on the condition y(1) = 1 is obtained as 2 3 2 3 8 ; 3 h i 7 e ; Fe ¼ 6 W 4 0 1 4 ; 3 5. 1 1 1 ; 1 Solving this system, the Taylor coefficients are obtained as a0 ¼ 1;
a1 ¼ 1;
a2 ¼ 1
3
7 0 2 5. 0 0
Hence the fundamental matrix equation becomes and the augmented matrix 2 2 3 8 6 ½W ; F ¼ 4 0 1 4
3
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1533
and thereby the polynomial solution becomes yðxÞ ¼ 1 þ x þ x2 . Note that from [W; F] we obtain the same result and taking N > 2 we can also obtain the same solution. Example 2. Let us now solve the problem y 00 ðxÞ xy 0 ðx 1Þ þ yðx 2Þ ¼ x2 2x þ 5; yð0Þ ¼ 1; y 0 ð1Þ ¼ 2; 2 6 x 6 0. We want to find the solution in the form yðxÞ ¼
3 X
an xn ;
n¼0
where c = 0, N = 3. The quantities in the given equation are P 00 ðxÞ ¼ 1; s00 ¼ 2;
P 10 ðxÞ ¼ x; s10 ¼ 1;
P 20 ðxÞ ¼ 1; s20 ¼ 0.
The matrix form of equation is 2 X 0 X 3 X k¼0
j¼0
pikj I i T ðskj ÞM k A ¼ F ;
i¼0
where p000 ¼ 1;
p100 ¼ 0;
p200 ¼ 0;
p010 ¼ 0;
p110 ¼ 1;
p020 ¼ 1;
p120 ¼ 0;
p300 ¼ 0;
p210 ¼ 0; p220 ¼ 0;
p310 ¼ 0; p320 ¼ 0
or simplifying I 0 T ðs00 Þ I 1 T ðs10 ÞM þ I 0 T ðs20 ÞM 2 A ¼ F ; where 3 0 07 7 7; 05 1 2 1 60 6 T ðs00 Þ ¼ T ð2Þ ¼ 6 40 0 2
1 60 6 I0 ¼ 6 40 0
0 1 0 0
0 0 1 0
T ðs20 Þ ¼ T ð0Þ ¼ I 0 ;
2
0 0 61 0 6 I1 ¼ 6 40 1 0 0 3 2 4 8 1 4 12 7 7 7; 0 1 6 5 0 0 1 F ¼ ½5
2
0 0 0 1
3 0 07 7 7; 05 0
2
1 0 0 0 1 60 6 T ðs10 Þ ¼ T ð1Þ ¼ 6 40 0
1
0T .
0 60 6 M ¼6 40 0 2
0 2 0 0 1 1 0 0
3 0 07 7 7; 35 0 3 1 1 2 3 7 7 7; 1 3 5 0 1
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Substituting these quantities in the for WA = F is obtained as 2 1 2 6 8 6 6 0 0 2 15 ½W ; F ¼ 6 6 0 0 1 0 4 0
0
0
2
previous equation, the augmented matrix 5
;
3
7 ; 2 7 7. ; 1 7 5 0
;
The augmented matrix for the given problem becomes 2 3 1 2 6 8 ; 5 7 h i 6 6 0 0 2 15 ; 2 7 7. e ; Fe ¼ 6 W 61 0 0 0 ; 1 7 4 5 0
1
2
3
; 2
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 a1 ðx þ 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 a1 ðx þ 2Þ þ x2 1. In addition, it is obtained the same results respectively both homogenous and non-homogenous cases for N P 2. Example 3. Let us consider the problem p pffiffiffi p
y 000 ðxÞ cosðxÞy 0 ðxÞ sinðxÞy 0 x þ 2y x ¼ sinðxÞ 2 cosðxÞ 1; 2 4 p y 00 ð0Þ ¼ 0; 6x60 yð0Þ ¼ 0; y 0 ð0Þ ¼ 1; 2 and approximate the solution yðxÞ ¼
5 X
an xn .
n¼0
The given equation is in the form 3 X 1 X k¼0
j¼0
P kj ðxÞy ðkÞ ðx skj Þ ¼ f ðxÞ;
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1535
where P 00 ðxÞ ¼ 0; s00 ¼ 0;
P 01 ðxÞ ¼
pffiffiffi p 2; s01 ¼ ; 4
P 20 ðxÞ ¼ 0; s20 ¼ 0;
p P 11 ðxÞ ¼ sin x; s11 ¼ ; 2 P 21 ðxÞ ¼ 0; s21 ¼ 0;
P 30 ðxÞ ¼ 1; s30 ¼ 0;
P 31 ðxÞ ¼ 0; s31 ¼ 0.
P 10 ðxÞ ¼ cos x; s10 ¼ 0;
The matrix form of this problem is 3 X 1 X 5 X k¼0
j¼0
pikj I i T ðskj ÞM k A ¼ F ;
i¼0
where pi00 ¼ pi20 ¼ pi21 ¼ pi31 ¼ 0 ði ¼ 0; 1; . . . ; 5Þ; pffiffiffi p001 ¼ 2; p101 ¼ p201 ¼ p301 ¼ p401 ¼ p501 ¼ 0; p110 ¼ 0;
1 p210 ¼ ; 2
p011 ¼ 0;
p111 ¼ 1;
p211 ¼ 0;
p030 ¼ 1;
p130 ¼ p230 ¼ p330 ¼ p430
p010 ¼ 1;
p310 ¼ 0;
p410 ¼
1 ; 24
1 p311 ¼ ; p411 ¼ 0; 6 5 ¼ p30 ¼ 0;
p510 ¼ 0;
p511 ¼
1 ; 120
T ðs00 Þ ¼ T ðs10 Þ ¼ T ðs20 Þ ¼ T ðs30 Þ ¼ T ðs21 Þ ¼ T ðs31 Þ ¼ I 0 ; 2
6 60 6 6 p 6 0 6 T ¼6 6 2 60 6 6 40 0 2
p2 4
p8
p4 16
1
p
3p2 4
p3 2
0
1
3p 2
3p2 2
0
0
1
2p
0
0
0
1
0
0
0
0
1 p2
1
6 60 6 p 6 60 T ¼6 6 4 60 6 6 40 0
3
3
5
p32
7 7 7 7 5p3 7 4 7 7; 5p2 7 7 2 7 7 5p 2 5 5p4 16
2
p2 16
p64
p4 256
5
1
p2
3p2 16
p16
5p4 256
0
1
3p 4
3p2 8
5p 32
p 1024
3
0
0
1
p
0
0
0
1
5p2 8 5p4
0
0
0
0
1
3 7 7 7 7 7 7; 7 7 7 7 5
3
3
7 6 6 1 7 7 6 7 6 6 1 7 7 F ¼6 6 1 7; 6 67 7 6 6 1 7 4 12 5 1 120
1
p4
3
3
2
0 1 0 0
6 60 6 60 6 M ¼6 60 6 6 40
0 2 0 0 0 3 0 0 0 0 0 0
0 0 0 0
0 0
3
7 0 07 7 0 07 7 7; 4 07 7 7 0 55 0 0
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2
1
6 60 6 60 6 I0 ¼ 6 60 6 60 4 0 2 0 6 60 6 61 6 I2 ¼ 6 60 6 60 4 0 2 0 6 60 6 60 6 I4 ¼ 6 60 6 61 4 0
0 0
0
0
1 0
0
0
0 1
0
0
0 0
1
0
0 0
0
1
0 0
0
0
0 0
0
0
0 0
0
0
0 0
0
0
1 0
0
0
0 1
0
0
0 0
1
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
3
2
7 07 7 07 7 7; 07 7 07 5 1 3 0 7 07 7 07 7 7; 07 7 07 5 0 3 0 7 07 7 07 7 7; 07 7 07 5 0
0
6 61 6 60 6 I1 ¼ 6 60 6 60 4 0 2 0 6 60 6 60 6 I3 ¼ 6 61 6 60 4 0 2 0 6 60 6 60 6 I5 ¼ 6 60 6 60 4 1
0
0
0 0
0
0
0 0
1
0
0 0
0
1
0 0
0
0
1 0
0
0
0 1
0
0
0 0
0
0
0 0
0
0
0 0
0
0
0 0
1
0
0 0
0
1
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
3
7 07 7 07 7 7; 07 7 07 5 0 3 0 7 07 7 07 7 7; 07 7 07 5 0 3 0 7 07 7 07 7 7. 07 7 07 5 0
Substituting these values in equation and simplifying, we have the fundamental matrix equation: pffiffiffi p
1 1 2T þ I 0 þ I 2 I 4 M 4 2 24
1 1 p 3 I5 T þ I 1 þ I 3 M þ M A ¼ F. 6 120 2 After the ordinary operations, the augmented matrix [W; F] is obtained, and augmented matrices for the conditions become ½U 1 ; k1 ¼ ½1
0
0
0
0 0
;
0;
½U 2 ; k2 ¼ ½0
1
0
0
0 0
;
1;
½U 3 ; k3 ¼ ½0
0
2
0
0 0
;
0.
e ; Fe is gained. Solving this Therefore the augmented matrix for the problem ½ W system, the approximate solution in terms of Taylor polynomial are obtained as yðxÞ ffi x 0:166712x3 þ 0:000118x4 þ 0:007794x5 .
M. Sezer, A. Akyu¨z-Dasßcıog˘lu / Appl. Math. Comput. 174 (2006) 1526–1538
1537
Table 1 Comparison of the absolute errors xi
N=5
N=8
N = 10
N = 12
N = 15
p 2 p 3 p 4 p 6
0.001523 0.000603 0.000192 0.000034 0
0.000031 0.000014 4.510809E6 7.929352E7 0
8.233735E6 4.927905E6 1.67317E6 3.092988E7 0
7.996475E7 7.107061E6 2.792385E6 5.73252E7 0
3.408115E7 6.002942E6 2.39564E6 4.969168E7 0
0
In addition, for N = 8 we have the Taylor coefficients A ¼ ½0 1
0
0:000195
0:166668
2:885E 6 0:008324
5:815E 6
6:56E 6
and we compare the absolute errors for different N in Table 1, since the exact solution of this problem is sin(x).
5. Conclusions The method presented in this study is useful in finding approximate and also exact solutions of certain differential–difference equations. In the cases m = 0 and skj = 0, the method can be used respectively for difference equation and differential equation. Differential–difference equations with variable coefficients are usually difficult to solve analytically. In this case, the presented method is required for the approximate solution. On the other hand, it is observed that the method has the best advantage when the known functions in equation can be expanded to Taylor series with converge rapidly (about x = c, skj 6 c 6 0). Also, the method can be used to solve the Fredholm integral and integrodifferential–difference equations. References [1] M. Sezer, A method for the approximate solution of the second order linear differential equations 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 ß . Nas, S. Yalc¸ınbasß, M. Sezer, A Taylor polynomial approach for solving high-order linear Fredholm integro-differential equations, Int. J. Math. Educ. Sci. Technol. 31 (2) (2000) 213– 225.
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[4] A. Akyu¨z, M. Sezer, A Chebyshev collocation method for the solution of linear integrodifferential equations, Int. J. Comput. Math. 72 (4) (1999) 491–507. [5] A. Akyu¨z, M. Sezer, Chebyshev polynomial solutions of systems of high-order linear differential equations with variable coefficients, Appl. Math. Comput. 144 (2003) 237–247. [6] T.L. Saaty, Modern Nonlinear Equations, Dover Publications, New York, 1981.