 Open Access
 Total Downloads : 404
 Authors : V. O. Ojo, R. B. Adeniyi
 Paper ID : IJERTV1IS9207
 Volume & Issue : Volume 01, Issue 09 (November 2012)
 Published (First Online): 02122012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
The Integrated Formulation of the Tau Method and Its Error Estimate For Third Order NonOverdetermined Differential Equations
Department of General Studies, OyoState College of Agriculture, IgboOra, Oyo State, Nigeria.
and
Department of Mathematics, University of Ilorin, Ilorin, Nigeria.
Corresponding Author
This paper is concerned with integrated formulation of the tau methods for numerical solution of initial value problems in nonoverdetermined third order ordinary differential equations. The error estimate for this variant of the tau method is obtained and numerical results are provided. The numerical evidence shows that the variant is more accurate than the differential variant earlier reported.
Accurate approximate solution of initial value problems and boundary value problems in linear ordinary differential equations with polynomial coefficients can be obtained by the tau method introduced by Lanczos in 1938. Techniques based on this method have been reported in literature with application to more general equation including nonliner ones as well as to both deferential and integral equations. We review briefly here two of the variants of the method.
Consider the following boundary value problem in the class of mth order ordinary differential equations:
m
r
p (x) y (r ) (x)
r 0
f (x), a x b
(1.1a)
rk
L* y(x )
m
ark
r 0
y (r ) (x )
k , k
1(1)m
(1.1b)
rk
where are given real numbers,
and the functions and
pr (x)
Nr
pr ,k
k 0
xk , r
0(1)m
.(1.2)
are polynomial functions or sufficiently close polynomial approximants of given real functions .
The number of overdetermination, s, of equation (1.1a) is defined as
(1.3)
for and .
Equation (1.1a) is said to be nonoverdetermined if s, given by (1.3) is zero, i.e. if s =
0. Otherwise it is overdetermined.
For the solution of (1.1) by the tau method ( Ortiz1969,1974, Lanczos 1938, 1956), we seek an approximant
yn (x)
n
r
a x r , n < + (1.4)
r 0
ofy(x) which satisfies exactly the perturbed problem
Lyn (x)
f (x)
m s 1
r 0
m s 1Tn m r 1 (x), a
x b , (1.5a)
n rk
L* y (x )
k , k
1(1)m
(1.5b)
where , r = 1(1)m +s, are fixed parameters to be determined along with , r = 0(1)n, in (1.4)by equating the coefficients of power of x in (1.5). The polynomial
Tr (x)
cos
r Cos 1 2x
b
2a 1
a
r
k
C (r ) xk
k 0
(1.6)
is the rth degree Chebyshev polynomial valid in [a, b] (see Fox and Parker 1968)
The integrated form of (1.1a) is given by
LY(x) =
m f (x)dx + (m(x) (1.7)
Where (m(x) denotes an arbitrary polynomial of degree (m 1), arising from the constants of integration, and
I L = m
L( )dx
(1.8)
is the m times indefinite integration of L (s). The integrated tau problem corresponding to (1.4) is therefore
I, ( y(x)) =
m f (x)dx + (m(x)) +
m s 1
m s 1Tn r 1 (x)
r 0
(1.9a)
n
L* (Y (xrk
) = k, k = 1(1)m .(1.9b)
where
n
Y (x) =
n
r
b xr
r 0
y(x), n + (1.10)
Problem (1.9) often gives a more accurate approximant of Y(x) than (1.4) does, due to its higher order perturbation term (see[7] and [14]).

We review briefly here error estimation of the tau method for the variants of the preceding section and which was earlier reported ( Adeniyi et al 1990, 1991 and 2007)

While the error function
en(x) = y(x) yn(x) (2.1)
satisfies the error problem ( see Onumanyi and Ortiz 1982)
m s 1
Len (x)
m s 1 Tn m r 1 (x)
r 0
(2.2b)
(2.2a)
The polynomial error approximant
(en(x))n+1 =
vm (x)
nTn m 1 (x)
(2.3)
C
( n m 1)
n m 1
ofen(x) satisfies the perturbed error problem ( Adeniyi et al 1990, 1991, Onumanyi et al 1882 )
L(en (x)) n 1
L* (e (x ))
m s 1
(
r 0
n rk n 1
0
m s 1 Tn m r 1 (x) +
m s 1 Tn m r
2 (x)) (2.4a)
(2.4b)
where the extra parameters
r r = 1(1)m + s, and in (2.3) (2.4) are to be
determined and in (2.3) is a specified polynomial of degree in which ensures that (en (x))n 1 satisfies the homogenous conditions (2.4b).
With 2.3) in (2.4) we get a linear system of m + s +1 equations, obtained by equating the coefficients of xn+s+1, xn+s, xn m +1, for the determination of by forward elimination, since we do not need the s in (2.3) consequently, we obtain an estimate
max (e (x)) =
n max (e (x)
(2.5)
C
a x b n n 1
(n m 1)
n m 1
a x b n

The error polynomial function
(e~ (x)) =
m (x) T
(x)
~ Y(x) y
(x) e (x)
(2.6)
C
n
n m 1
n (n m 1) n n
n m 1
Satisfies the perturbed integrated error problem
n n 1
I2 (e~ (x)) = m
m s 1
r 0
m s rTn m r 1 (x)dx
(m(x) +
m s 1
m s rTn r 1 (x)
r 0
(2.7a)
n
L* (e~ (xrk ))
= 0 (2.17b)
Equations (2.7) together with (2.6) yield a linear system of m + s +1 equations, obtained by equating the coefficients if xn + s + m + 1, xn+s+m, xn+1 for the determination of
n subsequently we obtain
max e~ (x)) n max e~ (x) = 4
C
n n 1
x x b
(n m 1)
n m 1
a x b n
We consider here the integrated form of the tau methods and its error estimate for the class of problems:
Ly(x): = ( 0 + 1x + 2×2 + 3×3 ) y (x) + ( 0 + 1x + 2×2 ) y (x)
+ ( 0 + 1x)y (x) + oy(x) =
n
r
f xr , a < x < b (3.1a)
r 0
y(a) = 0, y (a) = 1, y (a) = 2 (3.1b)
that is, the case when m = 3 and s = 0 in (1.1)
Without loss of generality, we shall assume that a = 0 and b = 1, since the
transformation
v (x
(b
a) , a < x < b (3.2)
a)
takes (3.1) into the closed interval [0, 1].
By applying (1.9) we have
x u t
0 0 0
( 0 + 1v + 2v2 + 3v3) y (v)dvdtdu +
x u t
0 0 0
( 0 + 1v + 2v2 + 3v3) y (v)dvdtdu
x u t
+
0 0 0
( 0 + 1v) y (v)dvdtdu + 0
x u t
0 0 0
y (v) dvdtdu
x u t F
=
f vr
dvdtdu + 1Tn + 3(x) + 2Tn + 1(x) = 3Tn+ 1(x)
r
0 0 0 r 0
That is
n n n n n
0 a rxr + 1 a rxr+1 + 2 a rxr+2 + 3 a rxr+3 0 0 3 1
arxr 1
r 0 r 0 r 0 r 0
r 0 r 1
n
6 2
arxr 2
n
9 3
arxr 3
+ x(2 1 0 0 1) + x
( 1 1 0 2 2 2 0)
2
r 0 r 2 r 0 r 3 2
+ 6 2
n
r 0 (r
arxr 2
1)(r 2)
+ 18 3
n
r 0 (r
arxr 3
2)(r 3)
n
6 3
arxr 3
n arxr 1 n
+ 0 + 1
arxr 2
n arxr 3
+ 2
( 0 0)u
r 0 (r
1)(r
2)(r 3)
r 0 r 1
r 0 r 2
r 0 r 3
2 2
n
r 0 (r
arxr 2
1)(r 2)
4 2
n
r 0 (r
arxr 3
2)(r
x2
+
3) 2
( 1 0 0 1)u
n
+ 2 2
arxr 3 n
+ 0
arxr 3
p>n arxr 3
+ 1
r 0 (r
1)(r
2)(r 3)
r 0 (r
1)(r
2)(r 3)
r 0 (r
2)(r 3)
x2 n
arxr 3
n arxr 3
2 ( 0
0 ) + 0
r 0 (r
1)(r
2)(r
1
3) r 0 (r
)(r
2)(r 3)
n f xr 3
n 3
( n 2) r
n 2
( n 2) r
n 1
( n 1) r
= r + 1
Cr x + 2
Cr x + 3
Cr x
r 0 (r 1)(r 2)(r 3) r 0
r 0 r 0
We equate corresponding coefficients of powers of x to obtain the system
0a0 1 C ( n 3)
2 C ( n 2)
3 C ( n 1) = 0 0 ( 0 2 1 + 0)a1 1 C (n 3)
0 0 0 1
1
2 C (n 2)
3 C (n 1) = 2 1 0 0 1 0 1
1
1 [(2 2 1 + 0)a0 +( 0 1]a1 + 2 0a2 2 1 C (n 3)
2 2 C (n 2)
2 2 2
2
2 3 C (n 1) = 1 0 0 2 2 2 0 0 0 + 1 0 1 1 [(k
3)1
k1
0 ]
a
k 1
[(k 2)2 3(k 2) 2) (k 3) 6(k 3) 6(k 2) ]+ 2 1 2 0 ak 2
(k 1)k
+ [(k 3)3 9(k 3)2 + 2(k 3) 18) 3 (k 3)2 + 7 (k 3 + 4) 2
+ (k 2) 1 + 0 0]ak 3
(k 2)(k 1)k
1 C ( n 3) + 2 C ( n 2) + 3 C ( n 1) =
f( k 3)
, k = 3(1)n … (3.11)
k k k
k (k
1)(k 2)
+ [1 (n 1)
(n
31 1)
0 ]
a
n
+ [(n 2)3 9(n 2)2 + 2(n 2) 18) 3 (n 2)2 + 7(n
n 1
2) + 4) 2 + (n 1) 1 + 0 + 0]a(n 2) 1 C ( n 3)
2 C ( n 2) + 3 C ( n 1) =
n(n 1)(n + 1)
fn 2
n 1 n 1
(n 1)(n)(n 1)
[n3 + 3n + 2) 2 + (n 1) 1 6n 2 + 0]an(n +2)(n + 1)
+ [(n 1)3 (n 1)2 + 2(n 1) 18) 3 ((n 1)2 + 7(n 1) + 4) 2]an 1
n(n +1)(n + 2)
n 2
1 C ( n 3)
n 2
+ 2 C ( n 2) =
(n
fn 1
2)(n
1)n
+ [(n3 9n2 + 2n 18] 3 (n2 + 7n + 4) 2 + (n + 1) 1 + 0 + 0]an
(n +1)(n + 2)(n +3)
3 C
=
( n 3)
n 3 (n
1)(n
fn
2)(n 3)
n
We solve this to subsequently obtain the approximant Y (x) .
For problem (3.1) we have, from (2.7)
x u t
0 0 0
( 0 + 1v + 2v2 + 3v3)(en(v)n + 1 dvdtdu +
x u t
0 0 0
( 0 + 1v + 2v2 )
(en(v)n + 1 dvdtdu +
x u t
0 0 0
( 0 + 1v )(en(v)n + 1 dvdtdu +
x u t
0 0 0
0 (en(v)n + 1 dvdtdu
x u t
= (
0 0 0
n
r
1
C(n)V r +
r 0
n 1
r
2
C ( n 1)V r +
r 0
n 2
r
3
C (n 2)V r ) dvdtdu
r 0
n 4
+ 1
C ( n 4) xr +
n 3
r
2
C ( n 3) xr +
n 2
r
3
C ( n 2) xr
r
r 0 r 0 r 0
From the coefficients if xn + 3, xn + 2, xn + 1, we get the system
C ( n 4) = [18 3(n +2)(1 (n +3)+10 2(n + 2)(n + 3)+6(2 2 6 3 + 0] C ( n 2)
1 n 4
( n 4)
( n 3)
n 2
6(n + 2)(n + 3)(n + 4)
C n
1Cn 3
+ 2Cn 3
1 n =
(n 1)(n 2)(n 3)
n 2
{12(n +5) 6 2(n +2)(n + 3) + 1(n + 5)((n + 2+ 4) 1(n +2)(n + 5)) C n 2
2(n + 2)(n + 3)(n + 5)
n 3
[9 3(n +1) (n +1)(n + 2)(18 3 + 10 2)) + 6(2 2 6 3 + 0 1) C ( n 2) }
6(n + 1)(n + 2) (n + 3)
C
+
( n 4)
1 n 2
( n 3)
C
+

n 2
( n 2)
C

n 2
C n
1 n 1
C ( n 1)
2 n 1
n(n 1)(n 2)
n(n 1)(n 2)
{[ 0
31 (n
21 2)
0 (n 2)
C +
n 2
n 2
(6 2(2(n + 5) (n +1)(n +2)) + ((n +1)(n + 5)( 1 1)) C n 2 ]
2(n +1)(n +2)(n + 5)
n 4
+ [(9 3n (1 2(n + 1) + 10 2n (n + 1) + 6(12 2 6 3 + 0 1)) C ( n 2)
6n(n +1)(n + 2)
C
(n 2)
3 n 1
n
C
1 n 2
C (n 1)
2 n 2
C (n 2)
3 n 2 =
n(n 1)(n 1)
(n 1)(n)(n 1)
(n 1)(n)(n 1)
{[ 62 (2
n) 1 (n
4) n1 C (n 2) + 0
31
21
0 C (n 2)
2n(n 1)
n 4 (n 2) n 3
n 5
+ 9 3(n 1)(n +2)(12n) +10 2n(n +1)(n 1) + 6 ((2 2 6 3 + 0 1) C ( n 2)
6n(n +1)(n 1) (n + 2) …(3.12)
n 2
where = n( C ( n 2) ) 1
n 2
n 2
we solve this system by forward elimination of n and subsequently obtain from (2.8) the error estimate
n 2
= 22n 4 3  12 C ( n )
+  2(n 1) C ( n 1)
2n C ( n 1)
(3.13)
2n(n + 1)(n 1)22n 4 p4
where
(6 (2 n) (n 4) n ) C (n 2)
(n 2) ( 3 )
p4 = 2 1 1 n 4
0 1 0
2n(n 1)
22n 5
2 (n 1)
n 5
+ (9 3(n 1)(n +2)(12n) +10 2n(n +1)(n 1) + 6 (n + 2)(2 2 6 3 + 0 1) C ( n 2)
6n(n + 1)(n 1) (n + 2) 22n 5
+ (n 2) p3]
2
p3 =
(0
31 (n
21 0 )
2)
(n 1)( (6 2(2n + 10) (n + 1)(n + 2) + (n + 1)(n + 5)( 2 1) 4(n + 1)(n + 2) (n + 5)
C (n 2)
+ (9 3n(2n 1) + 10 2n(n + 1) + 6 (2 2 6 3 + 0 + 1 n 5
22n 5
C (n 4) (n 3)
n 2 p1 + p2
22n 7 2
6n(n + 1) (n + 2)
p2 = 6(n +1)(n +2)(n +3)(12(n + 5)) 6 2( n +2)(n +3) + 1(n +5)((n + 2) + 4) 1((n
+ 2)(n + 5) (n 2) (18 2(n +1) (n +1)(n + 2)(18 2 10 2) + 3(2 2 6 3 + 0 1)
+ 3(n + 1)(n + 2) (n + 3)(n + 4)p1
6(n + 1) (n + 2) (n + 3)
p1 = 18 3(n + 2)(1 (n + 3) + 10 2(n + 2)(n + 3) + 6(2 2 6 3 + 0 1)
6(n + 2)(n +3)(n + 4)
In an earlier work ( Ojo and Adeniyi (2011), we obtained the corresponding error estimate for the differential form as
22 10n  
where
= 3
p7
… (3.14)
n 4
p7 = {[(n + 1)(n 1) 0 (n 1)(n 3) 2](n 1)(n 2) 1 + (n 1) 0) C ( n 2)
n 5
+ ((n 2)(n 3)(n 4) 3 + (n 2) 1 + 0) C ( n 2) n(n 1)(n 2)2 1
+ n(n 1)(n 2) 0]
Computed results from this are contained in Table 4.2 below.

A NUMERICAL EXPERIMENT
We consider here the following problem for experimentation with our results of the preceding sections. The exact error is defined by
* = max {y(xk)yn(xk)}, 0 < x < 1 {xk} = {0.01k}, for k = 0 (1)100
0 x 1
Example
Ly(x) = y (x) 5y (x) + 6y (x) = 0 , 0 < x < 1
y(0) = 0, y (0) = 1, y (0) = 0
The exact solution is
y(x) = 56
+ 3
2e2 x
2
3e3x
The numerical results are presented in Table 4.1 below.
Degree(n)
Error
5
6
7
~
9.88 x10 5
5.99 x 10 5
3.85 x10 5
*
1.96 x10 4
2.29 x10 5
4.25 x10 7

The integrated form of the tau method for the solution of Initial Value Problems (IVPs) in a class of third order differential equations with nonover determination has been presented. The error estimate is good as it closely captures the order of the error. This is better achieved than for the case of the direct series substitution approach otherwise referred to as the differential form thus lending credence to the preference of the former. This may be attributed to the higher order perturbation term which the integrated form involves.

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Ojo V.O and Adeniyi R. B. The Differential Form of the Tau Method and its Error Estimate for Third Order NonOverdetermined Differential Equations ( To Appear).
