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DERIVATION OF A CLASS OF HYBRID ADAMS MOULTON
METHOD WITH CONTINOUS COEFFICIENTS
Alagbe, Samson Adekola.1 Awogbemi, Clement Adeyeye2* Amakuro Okuata3
1, 3
Department of Mathematics, Bayelsa State College of Education, Okpoama, Nigeria.
2.
National Mathematical Centre, Sheda-Kwali, P.M.B 118, Abuja, Nigeria.
*
E-mail of the corresponding author: awogbemiadeyeye@yahoo.com
Abstract
This research work is focused on the derivation of both the continuous and discrete models of the hybrid
Adams Moulton method for step number k =1 and k = 2. These formulations incorporate both the off –
grid interpolation and off- grid collocation schemes. The convergence analysis reveals that derived
schemes are zero stable, of good order and error constants which by implication shows that the schemes
are consistent.
Keywords: Hybrids Schemes, Adams Methods, Linear K–Step Method, Consistency, Zero Stable
1.0 Introduction
The derivation of hybrid Adams Moulton Schemes of both the continuous and discrete forms for the off-
grid interpolation and the off – grid collocation system of polynomials is our primary focus in this
research work. Sequel to this would be to ascertain the zero stability of each of the discrete forms.
Performances of these schemes on solving some non stiff initial value problems shall be affirmed in the
second phase of this work which will focus on the application of these derived scheme and comparism of
the discrete schemes with the single Adams-Moulton Methods and its alternative in Awe (1997) and
Alagbe (1999)
In this paper we regard the Linear Multi-step Method and Trapezoidal ( Adams – Moulton Method) as
extrapolation and substitution methods respectively.
We also define k –Step hybrid schemes as follows:
k
y
i 0
i n i hk i f ni h r f nv . …………………………………………………(1.1)
i 0
where k 1, 0 and 0 are not both zero and v {0,1,...k} and of course f n v f ( xnv , ynv ) .
Gregg & Stetter (1964) satisfied the co-essential condition of zero-stability, the aim of which to reduce
some of the difficult inherent in the LMM (i.e poor stability) was achieved.
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Hybrid scheme was as a result of the desire to increase the order without increasing the step number and
then without reducing the stability interval. However, the hybrid methods have not yet gained the
popularity deserved due to the presence of off-grid point which requires special predicator which will not
alternate the accuracy of the corrector to estimate them.
1.1 Adams Methods
This is an important class of linear multi-step method of the form:
1 5 3 251 4
y n1 y n h 1 2 3 .. . f n ................................(1.2)
2 12 8 720
The corrected form of Adams-Moulton form is expressed as:
1 1 1 19 4
y n1 y n h 1 2 3 ... f n 1 ................................(1.3)
2 12 24 120
The trapezoidal scheme is a special case of the Adams –Moulton method, in which only the first two terms
in the bracket is retained. This method is of the highest order among the single –step method. It is being
expressed as
y n 1
h
f n1 f n
2
2.0 Derivation of Continuous and Discrete Hybrid Adams-Moulton Scheme
Our concern primarily is to derive the continuous and discrete form of both one-step and two-step
Adams-Moulton scheme and consequently carried out the error and zero-stable analysis of the discrete
forms. Matrix inversion technique was the tool implored to achieve the derivation of the continuous form.
2.1 Derivation of Multi-Step Collocation Method
In order to derive the continuous and discrete one and two step hybrid Adams-Moulton schemes, we
employed the approach used by Sirisena (1997) where a k-step multi-step collocation method point was
obtained as:
k 1 m 1
y x j x y x n j h j x f x j , y x j .......................................................2.1
j 0 j 0
Where j x and j x are the continuous coefficients
We defined j x and j x respectively as:
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t m 1 i
j x j ,i 1 x
i 0
t m 1
h j x h j ,i 1 x i ...............................................................................................2.2
i 0
To get j x and j x we arrived at matrix equation of the form DC=I………… (2.3)
,
where I is the identity matrix of dimension (t + m) (t + m), D and C are defined as
1 xn xn 2
xn t
x n m 1
t
1 x n 1
2
x n 1 t
x n 1 x n 1 1
t m
.
2 t t m 1
1 x n k 1 x n k 1 x n k 1 x n k 1
D _
.........................................2.4
t 1 t m 2
0 1 2 xo t x0 t m 1 x 0
t 1 t m 2
1
0 2 x m 1 t x m 1 t m 1 x m 1
Thus, matrix (2.4) is the multi-step collocation matrix of dimension (t+m) (t+m) while matrix C of the
same dimension whose columns give the continuous coefficients given as:
01 11 t 1,1 h 01 h m1,1
12 t 1, 2 h 02 h m1, 2
C
02 ..........................2.5
0 j m
1 j m t 1, j m h 0 j m h m1,m j
We define t as the number of interpolation points and m is the number of collocation points. From (2.3),
we notice that If we define
This implies that a suitable algorithm for getting the elements of C is the following:
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C ij eij u i ,k c k , j , i j 1,2,...n,.........................................................................................2.6
ei , j
ei , j , j 1,2,...n.
li , j
i 1
eij eij lik ekj lij , i 1,2,...n., j 1,2,...n...........................................................................2.7
1
k 1
i 1
lij d ij lik u kj , j i, j 1,2,...n.; i 1,2,..., n
k 1
lij d ij , i 1,2,..., n........................................................................................................................2.8
i 1
u ij d ij lik u kj lii , i, j , i 1,2,..n.
k 1
u ij d ij lii ; j 1,2,...n,..................................................................................................................2.9
Provided
2.2 Derivation of Continuous and Discrete Hybrid Adams-Moulton Scheme
Case K=1
The matrix for this case is given below as
1 x n xn2
xn
3
2 3
1 x n u x n u x n u
D
0 1 2 xn 3x n and its equation (2.1) equivalence is
0 1
2 x n 1 3 x n 1
yx 0 x y n u x y nu h 0 x f n 1 x f n1
By applying the set of formulae (2.6) - (2.9, we have
li1 d i1 , i 1,2,3,4,....................................................................................(2.10)
l11 d11 1
l 21 d 21 1
l31 d 31 0
l 41 d 41 0
Now using
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u ij d ij lij , j 1,2,3,4, i 1
j2
d12
u12 xn
l11
j3
d13
u13 xn
2
l11
j4
d14
u14 xn
3
l11
Using
j 1
l ij d ij lik u kj , ij
k 1
i 2, j 2
l 22 d 22 lik u k 2
k 1
l 22 d 22 l 21u12
l 22 x n uh x n uh
i 3, j 2
l 32 d 32 l31u12
l 32 1 0.x n 1, i 4, j 2
l 42 d 42 l 41u12 1 0, x n 1
Using
i 1
uij d ij lik ukj lii , i j, i 1,2,
k 1
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i 1, j 4
u
d
23
l u
21 13
23 l
22
x uh 2 x .1
n n
uh
2 x uh
n
i 2, j 4
u2 4
d 2 4 l2 1u1 4
l2 2
x n uh x 3 .1
3
uh
3 xn
2
3 xn uh u 2 h 2
Again using
i 1
l ij d ij l ik u kj , i j
k 1
i 3, j 3
l 33 d 33 l31u13 l32u 23
2 x n 2 x n uh uh
i 4, j 3
l 43 d 43 l 41u13 l32u 23
2 x n 2h 2 x n uh
2 u h
Using
i 1
u ij d ij lik u kj lii , i j , i 3, j 4,
k 1
d l31u14 l32u 24
u 34 34
l33
3x 3 x n 3 x n uh u 2 h 2
2
n
2
uh
3 x n uh
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Using
i 1
lij d ij lik u kj , j i, i 4, j 1
k 1
l 44 d 44 l 41u14 l 42u 24 l 43u 34
3 x n h 3x n 3x n uh u 2 h 2 2h uh 3x n uh
2 2
3h 2 2uh 2 3 2u h 2
Using (2.7) with
i j 1
e11
eij e11
1 1
1
l11
j2
e12
e12
1
0
l11
j3
e13
e13
1
0
l11
j4
e14
e14
1
0
l11
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i 1
eij eij l ik ekj l ij , i 1,2,3,4; j 1,2,3,4.
1
k 1
i 2, j 1
e l e
e1 21 21 11
21
l 22
0 1 1
uh uh
e31 31
1
e l31e11 l 32e1
1
21
l 33
1 1 1
2 2
uh uh u h
i 4, j 1
e 1
e 41
l 41e11 l 42e1 l 43e31
1
21
1
41
l 44
1 1
2h uh 2 2
u h
uh
3 2u h 2
2
3 2u u 2 h 3
i= 2=j
e1
e 22 l 21e12
1
22
l 22
1
uh
i 3, j 2
e32
1 e 32
l31e12 l32e1
1
22
l33
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1 1 1
2 2
uh uh u h
i 4, j 2
e1
e 42
l 41e12 l 42e1 l 43e32
1
22
1
42
l 44
1 2 u h
2 2
u h
uh
3 2u h 2
2
3 2u u 2 h 2
i 2, j 3
e1
e 23
l 21e13
1
23
l 22
0 1.0 0
l 22
i 2, j 4
e 1
e 24 l 21e14
1
24
l 22
0 1.0 0
uh
i j 3
e33
1 e l
33 e l32e1
1
31 13 23
l33
1
uh
i 4, j 3
e1
e 43
l 41e13 l 42e1 l 43e33
1
23
1
43
l33
2 u
uh
3 2u h 2
2 u
3 2u uh 2
i 3, j 4
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e34
1 e 34
l31e14 l 32e1
1
24
l33
0
0
uh
i j4
e1
e 44
l 41e14 l 42e1 l 43e34
1
24
1
44
l 44
1
3 2u h 2
C values are now computed thus;
2
c 41 e1
u h 3 2u
41 2 2
i 4, j 2
2
c 42 e1
u h 3 2u
42 2 2
i 4, j 3
c 43 e1
2 u
uh 2 3 2u
43
i j4
1
c 44 e1
44
3 2u h 2
Using
cij eij u ik c kj
1
k 4
i 3, j 1
c31 e31 u 34c 41
1
1 2
2 2
3 x n uh 2 2
u h 3 2u
u h
6 xn 3h
u 2 h 3 3 2u
i 3, j 2
c32 e32 u 34c 42
1
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1 23 x uh
2 3n
u h22
u h 3 2u
3h 2 x n
2 3
u h 3 2u
i j3
c33 e33 u 34c 42
1
1 2 u
3 x n uh 2
uh uh 3 2u
u
3 h 3u 2 x n
2
uh 2 3 2u
i 3, j 4
c34 e34 u 34c 44
1
3 x n uh
3 2u h 2
Now with
k 1
cij eij u ik c kj
1
k 3
i 2, j 1
c 21 e1 u 23c 21 u 24c 41
21
2
1 2 x n uh 6 x n 3h 3 x n 3 x n uh 2u 2 h 2
uh u 2 h 3 3 2u u 2 h 3 3 2u
6x x
2 3 n n 1
u h 3 2u
i2 j
c 22 e1 u 23c32 u 24c 42
21
1 3h 2 x n 2
2 x n uh 2 3
2 3 3x x u 2 h 2
uh 3 2 xn u h u h 3 2u n nu
6 x n x n 1
u h 2 3 2u
2
i 2, j 3
c 23 e1 u 23c33 c 43u 24
23
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2 x n uh
u 2 h 3ux n 6 x n 3h 2 2 2 u
3 x n 3 x n uh u h 2
2
uh 3 2u
2
uh 3 2u
6 x n x n 1 3u x n h 2 u 2 h x n 2h
2
uh 2 3 2u
i 2, j 4
c 24 e1 u 23c34 u 24c 44
24
3xn uh
2 x n uh
1
3 x n uh u 2 h 2
3 2u h 2
3 2u h 2
x 3 x 2uh
n 2 n
h 3 2u
Now consider
k 2
Cij eij u ik c kj
1
k 2
i 1 j
c11 e11 u12c 21 u13c31 u14c 41
1
6x x 2 32 x h 3 2
1 x n 2 3 n n 1 x n 2 3 n xn 2 3
u h 3 2u u h 3 2u u h 3 2u
3u 2 h 3 2u 3 h 3 3 x n h 2 x n
2 3
u 2 h 3 3 2u
i 1, j 2
c12 e12 u12c 22 u13c34 u14c 42
1
xn 6 x 1
2
n
6x x 3 2
x n 2 3 n n 1 2 2
xn 2 3
u h 3 2u u h 3 2u u h 3 2u
3
3x h 2 x
2
n
3
n
u h 3 2u
2 3
i 1, j 3
c13 e13 u12c 23 u13c33 u14c 43
1
xn
6 x 2
3ux n 6 x n h u 2 x n h 2u 2 h 2 3uh 2
2
x u 2 h 3ux n 6 x n 3h
n 3
xn 2
2 u
n
uh 3 2u
2
uh 3 2u
2
uh 3 2u
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3x h u 2
n
2
x n h ux n 2u 2 x n h 2 3ux n h 2 2 x n
2 3 3
uh 2 3 2u
i 1, j 4
c14 e14 u12c 24 u13c34 u14c 44
1
x n 3x n 2ux n h
2
x n 3x n uh
2
2
3
xn
h 2 3 2u h 2 3 2u h 3 2u
x n x n u
2
h 2 3 2u
The continuous scheme coefficients column by column for c values are now computed.
Substituting these values into the general form (2.10) yield the desired continuous schemes.
0 x c11 c 21 x c 31 x 2 c 41 x 3
3u 2
h 2 2u 3 h 3 3 x n h 2 x n
2 3
6 x 2
n 6xn h x 6 x
n 3h x 2
2x 3
u 2 h 3 3 2u u 2 h 3 3 2u u 2 h 3 2u u 2 h 3 3 2u
3
2x x n 3 3hx x n 2 u 2 h 3 3 2u
u 2 h 3 3 2u
1 x c12 c 22 x c 23 x 2 c 24 x 3
3x 2
n h 2xn
3
6 x h 6 x x 6 x
n
2
n n 3h x 2
2x 2
u 2h3 u 2 h 3 3 2u u h 3 2u u 2 h 3 3 2u
2 3
3h x x n
2
2 x x n
3
u 2 h 3 3 2u
0 x c13 c 23 x c 33 x 2 c 34 x 3
3x 2
n h 2u 2 x n h 3u 2 x n h 2 2u 2 x n h 3ux n h 2 x n
2 2 2 3
uh3 2u
x 3
n 6 x n 3ux n 6 x n h 3u 2 x n h u 2 x n h 2u 2 h 2 3uh x
2 2
uh 3 2u
2
u 2
h 2 3ux n 6 x n 3h 2 u x 3
uh 2 3 2u uh 2 3 2u
2 u x x n 3 3h u 2 h x x n 2 2u 2 3u x x n h
uh 2 3 2u
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1 x c14 c 24 x c34 x 2 c 44 x 3
ux h x 3x
2
n
3
n
2
n
2
2ux n h x 3x n uh x 2
2
x3
h 3 2u
2
h 3 2u
2
h 3 2u h 3 2u
uh x x n
2
x xn 2
3
h 3 2u
Hence forth,
2x x n 3 3h x x n 2 u 2 h 3 3 2u
2x x n 3 3hx x n 2 .
yx yn yn u
u h 3 2u
2 3
u h 3 2u
2 3
3 2
2
2 u x x n 3 u hx x n 2u 3u h x x n
2 2
x x n uhx x n 2
3
fn f n 1 ....2.11
uh 3 2u
2
h 3 2u
2
Where (2.11) is the continuous form of one step Adams-Moulton scheme for k=1. If
1
(2.11) is evaluated at x n 1 and a substitution u is made, the result is
2
yxn 1 y n 2 y n 12
h h
f n f n1, hence
4 4
y n1 y n 2 y n 12
h
f n1 f n ........................................................................................................2.12
4
If on the other hand, equation (2.11) is differentiated with respect to x and then evaluated at
x xn 12 the result obtained is
,
y x h y
1 3 3 5 1
n 12 n y n 12 f n f n 1,
h 8 8
yn y
n
1
h
24
f n 1 5 f n 8 f n 12 ...........................................................................2.12b
2
The schemes (2.12a) and (2.12b) can be referred to as (2.12) and are indeed of interpolation
polynomial of case k =1
We consider another system of matrix of the same case k=1 where the schemes shall be derived
at the interpolation points
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1 xn xn2
xn
3
0 1 2 xn 3xn
2
D
0 1 2 x n 1 3x n 1
2
0 1 3x n u
2
2 x n u
The general forms of this system of matrix is given as :
yx 0 x y n h 0 x f n 1 x f n1 n x f nu ............................................................2.13
As seen in the first case using the same set of formulae (2.6) – (2.9), we have the values of c given as
follows:
c 41 0;
c 42
u 1
3uh 2 u 1
u
c 43
3uh u 1
2
1
c 44
3uh u 1
2
c31 0
2 x n uh h
c32
2uh 2
uh 2 xn
c33
2h 2 u 1
2 x n 12
c34
2uh 2 u 1
c 21 0
c 22
uh x n ux n h x n h
22
uh 2
c 23
ux n h x n2
h 2 u 1
x x
c 24 2n n 1
uh u 1
c11 1
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6ux n h 2 3ux n h 3x n h 2 x n
2 2 3
c12
6uh 2
3x n h 2 x n
2 3
c13
6uh 2 x 1
Following the same procedure as in the first case, the coefficients for the continuous schemes,
j x and j x are obtained and by similar substitution into equation (2.13), the desired continuous scheme
is obtained as follows;
y x 1y n
2u 1x x n 3 3h u 2 1 x x n 2 6uh 2 u 1x x n
fn
6uh u 1
2
2u x x n 3u hx x n
2 x x n 3hx x n
3 2 2 3 2
f n 1 f n u ..............................2.14
6uh u 1
2
6uh u 1
2
Putting u = ½ and proceeding to get our discrete forms as did earlier by evaluating (2.14) at these two
different points x xn1 and
xn xn1 , we have:
a. at x = xn = xn+1, the result is
h h 4
y n 1 y n f n f n 1 hf n 12
6 6 6
y n 1 y n f n f n 1 4 f n 12 ...........................................................................2.15a
h
6
b. at x = x + xn+ ½ , the result is
y n 12 y n
h
24
5 f n f n1 8 f n 12 ......................................................................2.15b
The collocation schemes of D yields just the same equation in (2.15b) . These are the hybrid collocation
schemes of the case k=1
Equation (2.15a) and (2.15b) can be referred to as (2.15)
We also considered two different systems of matrices D3 and D4 on the derivation of both hybrid
interpolation and the collocation where k = 2
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1 x n 1 2
x n 1 3
x n 1 x n 1
4
2 3
0 1 2 xn 3xn 4 xn
D3 0 1 2 x n 1 2
3x n 1 4 x n 1
3
2 3
0 1 2 xn2 3xn 2 4 xn2
0 1 2
4 x n u
3
2 x n u 3x n u
The general forms of D3 is given as
yx 1 x y n1 h 0 x f n 1 x f n1 2 x f n 2 n x f nu
Appling the set of formulae (2.6)-(2.9) on this system gives the c-entries as follows
c51 c 41 c31 0, c11 1
1
c52
8uh 3
1
c53
4uh u 2 u 1
3
4uh 12 x n 12h
c 42
24uh 3
x n 42 u h
12
c 43
12h 3 u 1
3x n uh h
c 44
6h 3 u 2
x n 1
c 45
uh u 2 u 1
3
c32
2uh 2 2uh 2 6 x n h 2h 2 2ux n h
4uh 3
c33
2uh 2 2ux n h 3 x n 4 x n h
2
2h 3 u 1
c34
3x n 2 x n h 3x n 4 x n h
2 2
4h 3 u 2
c35
3x 6 x n h 2h 2
2
n
2uh 3 u 2 u 1
c 22
2uh 3
x n 3 x n h 3ux n h 2 ux n h 2 x
3 2 2
2uh 3
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c 23
2ux n h 2 ux n h 2 x n h x n
2 3
h 3 u 1
c 25
2
3x n h 2 x n h 2 x n
3
uh 3 u 1
c12
24ux n h 3 18ux n h 2 4ux n h 3 x n 12 x n h 2 12 x n h 2 10uxh 4 3h 4
2 3 4 3 2
24uh 3
c13
3 x n 5h 4 8uxh 4 12ux n h 2 4ux n h 8 x n h
2 3 3
12u 1h 3
c14
4 x n h 4ux n h 6ux n h 2 3 x n 2uh 4 h 4
3 3 2 4
24h 3 u 2
c15
4 xn h 4 xn h 2 xn h 4
3 2 4
4uh 3 u 2 u 1
As in the previous cases, continuous schemes coefficients i x ' s and the i x ' s are obtained and
substituted into the general form (2.16) to obtain the desired continuous schemes as follows;
y x 1y n 1
3x x 4u 12 x x n h 18u 12 x x n h 2 24uh 3 x x n 3 10u h 4 f n
n
4 3 2
24uh 3
3x x n
4
42 u h x x n 12uh 2 x x n 5 8u h 4 f n 1
3 2
12h 3 u 1
3x x n
4
4hu 1 x x n 6uh 2 x x n 2u 1h 4 f n 2
3 2
24h 3 u 2
x x n 4x x n h 4x x n h 2 h 4 f n u
4 3 2
.............................................................................2.17
4uh 3 u 1u 2
Evaluating (2.17) at x x n 2 u 3 2 the resulting scheme is given as:
,
y n 2 y n1
h
6
f n1 f n 2 4 f n 3 2 ....................................................................................................2.18a
If (2.17) is evaluated at points x x
n u and u
3
2 , we have
y n 3 2 y n1
h
192
f n 46 f n1 5 f n 2 56 f n 3 2 .............................................................................2.18b
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3
However, if the point u 2 is maintained and (2.17) is evaluated at x = xn , the result is another discrete
scheme of the form:
y n1 y n
h
6
2 f n 7 f n1 f n 2 4 f n 3 2 .........................................................................................2.18c
Equations (2.18a), (2.18b) and (2.18c) are referred to as (2.18)
Taking on another system of matrix D4 of an interpolation scheme, where k = 2:
1 x n 1 2
x n ! 3
x n 1 x n 1
4
2 3 4
1 x n u x n u x n u x n u
D 4 0 1 2 xn 3xn 2
4 xn
3
2 3
0 1 2 x n 1 3x n 1 4 x n 1
0 1 2
4 xn2
3
2 xn2 3xn 2
the general form of the matrix is given as:
yx 1 x y n1 u x y nu h 0 x f n n x f n1 2 x f n 2 .......................................2.19
As in the previous cases, matrix C is determined with the columns simplification yielding the continuous
schemes:
yx
x x 4 2x xn1 2 h 2 u 12 u 2 2u 1h 4 y nu
n 1
u 12 u 2 2u 1h 4
x xn1 4 2x xn1 2 h 2 yn1 2u 5x x 4 2u 2 2u 1h 3 f
u 12 u 2 2u 1h 4 n 1 n
u 2 2u 2x xn1 4 u 1u 2 2u 1x xn1 3h 2u 2 2u 2h 3 f n1
3u 1u 2 2u 1h 3
2u 1x xn1 4 2u 2 2u 1x xn1 3 h u 13u 1x xn1 2 h 2 f n2 ..............2.20
12 u 2 2u 1 h 3
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Desired hybrid point of elevation remains U 3 and so (2.20) is evaluated to yield
2
7 y n 2 9 y n1 16 y n 3 2
h
f n 32 f n1 19 f n2 ................................................................2.21a
12
at the point x x
n 2
If consideration at the point x x is made, we have:
n
9 y n1 7 y n 16 y n 3 2
h
9 f n 48 f n1 3 f n3 ........................................................................2.21b
4
Finding the differential expression with respect to x for (2.20) and evaluating the differential coefficient at
xx the result becomes:
n 1 ,
y n1 y n 3 2
h
192
f n 46 f n1 5 f n 2 56 f n 3 2 ..........................................................................2.21c
As usual, equation (2.21a),(2.21b) and (2.21c) are referred to as (2.21)
3.0 Convergence Analysis
This section shows the validity and consistency of the derived scheme in section two . The tools for the
assignment would be the familiar investigation of the zero stability by finding the order and error constant
of the each of the schemes.
3.1 Definitions
k
( a) The scheme (1.1) is said to be zero stable if no root of the polynomial p d i has modulo
i
i 0
greater than one and every root with modulo one must be distinct or simple.
( b) The order p and error constant c p 1 for (1.1) could be defined thus:
If c c ...c 0, c then the principal local term error at x n k is c h p 1 y p 1 x .
0 1 p p 1 0 , p 1 n
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cq
1
q!
t q 0 1 t q 1 2 t q 2 ... k t q
1
q 1!
t q 1 0 1 t q 1 1 2 t q 1 2 ... k t q 1 k v t q 1 v ......................3.1
(c) A numerical method (1.1) is said to be consistent if p 1, where p is the order of the method.
3.1 Example
We take on the derived schemes in section 2 above, showing that each of the schemes is in conformity to
the definition above or otherwise.
Example 3.1.1
One of the derived schemes for step number k=1 is given as:
y n1 y n 2 y n 12
h
f n1 f n .........................................................................................2.12a
4
From equation (2.12a)
1 1
0 1, 1 1, 2, 0
1 , 1
2
4 4
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c0 12 1 1 0;
1
c1 12 0 1 12
2
1 12
2 24
0;
1 1 1
2
1 1 1 1
c 2 ! 1 1 ( ) 0
2 2 2 2 8 24 3 2
1
3
1 1 1 1 1 1 1 1 1 1 1 1
c3 ! 1 1 ( ) 2 1
0
3 2 2 2 2 2
48 2 24 12 48 2 24 48 48
1
4 3
1 1 1 1 1 1 1 1 1
c 4 ! 1
4 2 2 3! 2 2 384 144 144 384
1
So that the order p = 3 and c p 1 ; we seek the root of the characteristics polynomial as follows:
384
p 0 12 1 0
1
2
1
Hence, by definitions 3.1, we concluded that the method is zero stable.
We consider the scheme y n 3 2 y n 1
h
192
f n 46 f n1 5 f n 2 56 f n 3 2
1 5 23 7
Here, 1 1, 3 2 1, 0 , 2 , 1 , and , 3 2
192 192 96 24
Then observe that c0 c1 ..................................................................................c4 0
However,
c5
1
5!
1 3 2 5 3 2 1 2 4 2 3 2 4 3 2
1
4!
1 243 1 46 80 4236
1
120 32 24 192 192 3072
1 211 3692
3072
120 32 24
4.87 10 3
The order p=4 and c p 1 4.87 10 3. the absolute root for the polynomial is thus computed;
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Vol.2, No.10, 2012
p 3 2 1 0
3 3
2 2
i.e
1
2
1 0
0, or , 1, and , so, 1
Hence, the method is zero stable.
4.0 Conclusion
There is no doubt that the schemes are consistent and zero stable and could also be used by Numerical
Analysts to solve differential equations experimentally. The obtained result in comparism with the
theoretical result would also be of great importance in future research work.
References
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Mathematics, University of Jos,
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B.Sc Project, Department of Mathematics, University of Jos
Ciarlet, P.G. & Lions, J.L (1989). “Finite Element Method` Handbook of Numerical Analysis”. Vol.11
Fatunla S. O. (1988), “Computer Science and Scientific Computational Method for IVP in ODE”.
Academic Press Inc.
Gragg,W.B & Stetter H.J (1964), “Generalized Multistep Predictor-Corrector Methods”.
J-Assoc.Comp.Vol11, pp188-200,MR 28 Comp 4680
James, R. e tal (1995), “The Mathematics of Numerical Analysis”. Lecture in
Applied Mathematics.Vol XXXII.
Kopal, Z.(1956), “Numerial Analysis”. Chapman and Hall, London.
Lapidus L.B & Schiesser W.E. (1976), “Numerical Methods for Differential System”.
Academic Press Inc.
Onumanyi, P. et al (1994), “New Linear Multistep Methods for First Order Ordinary Initial Value
Problems“. Jour. Nig. Maths Society, 13 pp37 – 52
Sirisena, U.W. (1997), “A Reformation of the Continous General Linear Multistep Method by Matrix
Inversion for First Order Initial Value Problems”. Ph.D Thesis, Department of Mathematics, University of
Ilorin, Ilorin (Unpublished)
44
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