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Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.10, 2012
21
DERIVATION OF A CLASS OF HYBRID ADAMS MOULTON
METHOD WITH CONTINOUS COEFFICIENTS
Alagbe, Samson Adekola.1 Awogbemi, Clement Adeyeye
2* Amakuro Okuata
3
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
i i
vnriniini fhfhky0 0
. …………………………………………………(1.1)
where 00,1 andk are not both zero and v },...1,0{ k and of course ),( vnvnvn yxff .
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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ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.10, 2012
22
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:
)2.1..(.................................720
251
8
3
12
5
2
11 432
1 nnn fhyy
The corrected form of Adams-Moulton form is expressed as:
)3.1..(.................................120
19
24
1
12
1
2
11 1
432
1
nnn fhyy
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
nnn ffh
y 112
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:
1.2.......................................................,1
0
1
0
m
j
jjjn
k
j
j xyjfxhxyxxy x
Where xj and xj are the continuous coefficients
We defined xj and xj respectively as:
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ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.10, 2012
23
1
0
1,
1
0
1,
2.2...............................................................................................mt
i
i
ijj
mt
i
i
ijj
xhxh
xx
To get xj and xj, 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
4.2.........................................
1201
1210
1
.
1
1
2
1
1
11
_2
0
1
0
1
11
2
11
1
11
2
11
12
mt
m
t
mm
mtt
o
mt
kn
t
knknkn
mt
n
t
nnn
mt
n
t
nnn
xxx
xx
mtt
mttx
xxxx
xxxx
xxxx
D
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:
5.2..........................
,10,110
2,1022,11202
1,1011,11101
jmmmjmjtmjmj
mt
mt
hh
hh
hh
C
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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ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.10, 2012
24
9.2.........................................................................................................,.........,...2,1;
.,..2,1,,,
8.2........................................................................................................................,...,2,1
,...,2,1.;,...2,1,
7.2...........................................................................,...2,1.,,...2,1,
.,...2,1,
6.2................................................................................,.........,...2,1,
1
1
,
1
1
,
1
1
1
,
,
,
,,
njldu
nijiluldu
nidl
ninjijuldl
njnilelee
njl
ee
njicueC
iiijij
ii
i
k
kjikijij
ijij
i
k
kjikijij
ij
i
k
kjikijij
ji
ji
ji
jkkiijij
Provided
2.2 Derivation of Continuous and Discrete Hybrid Adams-Moulton Scheme
Case K=1
The matrix for this case is given below as
11
32
32
3210
3210
1
1
nn
nn
ununun
nnn
xx
xx
xxx
xxx
D and its equation (2.1) equivalence is
1100
nnunun fxfxhyxyxxy
By applying the set of formulae (2.6) - (2.9, we have
0
0
1
1
)10.2.....(......................................................................,.........4,3,2,1,
4141
3131
2121
1111
11
dl
dl
dl
dl
idl ii
Now using
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25
3
11
1414
2
11
13
13
11
1212
4
3
2
1,4,3,2,1,
n
n
n
ijijij
xl
du
j
xl
du
j
xl
du
j
ijldu
Using
1,01
2,4,1.01
2,3
2,2
,
12414242
32
12313232
22
12212222
1
22222
1
1
n
n
nn
k
kik
j
k
kjikijij
xuldl
jixl
uldl
ji
uhxuhxl
uldl
uldl
ji
ijuldl
Using
,2,1,,1
1
ijiluldu ii
i
k
kjikijij
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26
222
33
22
14212424
33
1.
4,2
2
1.2
22
13212323
4,1
huuhxx
uh
xuhx
l
uldu
ji
uhn
x
uh
nxuh
nx
l
uldu
ji
nn
n
Again using
hu
uhxhx
ululdl
ji
uhuhxx
ululdl
ji
jiuldl
nn
nn
i
k
kjikijij
2
222
3,4
22
3,3
233213414343
233213313333
1
1
,
Using
uhx
uh
huuhxxx
l
ululdu
jijiluldu
n
nnn
ii
i
k
kjikijij
3
333
,4,3,,
2222
33
2432143134
34
1
1
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27
Using
222
2222
3443244214414444
1
1
,
2323
32333
1,4,
huuhh
uhxuhhhuuhxxhx
ulululdl
jiijuldl
nnnn
i
k
kjikijij
Using (2.7) with
0
4
0
3
0
2
1
1
11
141
14
11
131
13
11
121
12
11
111
11
1
l
ee
j
l
ee
j
l
ee
j
l
eee
ji
ij
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28
44
1
3143
1
2142
1
1141411
41
22
33
1
2132
1
1131311
31
22
1121211
21
1
1
1
1,4
111
110
1,2
.4,3,2,1;4,3,2,1,
l
elelelee
ji
huuhuh
l
elelee
uhuh
l
elee
ji
jilelee ij
i
k
kjikijij
32
2
22
23
2
23
12
1
huu
hu
huuhh
uh
i = 2 = j
33
1
2232
1
1231321
32
22
1
1221221
22
2,3
1
l
elelee
ji
uh
l
elee
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29
4,3
23
2
23
2
3,4
1
3
00.10
4,2
00.10
3,2
23
2
23
21
2,4
111
2
2
33
1
3343
1
2342
1
1341431
43
33
1
2332
1
1331331
33
22
1
1421241
24
22
22
1
1321231
23
22
2
22
44
1
3243
1
2242
1
1241421
42
22
ji
uhu
u
hu
uh
u
l
elelelee
ji
uh
l
elelee
ji
uh
l
elee
ji
l
l
elee
ji
huu
hu
hu
hu
uh
l
elelelee
ji
huuhuh
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30
2
44
1
3443
1
2442
1
1441441
44
33
1
2432
1
1431341
34
23
1
4
00
hu
l
elelelee
ji
uh
l
elelee
C values are now computed thus;
2
1
4444
2
1
4343
22
1
4242
22
1
4141
23
1
4
23
2
3,4
23
2
2,4
23
2
huec
ji
uuh
uec
ji
uhuec
ji
uhuec
Using
4234
1
3232
32
2222
4134
1
3131
4
1
2,3
23
36
23
23
1
1,3
cuec
ji
uhu
hx
uhuuhx
hu
cuec
ji
cuec
n
n
k
kjikijij
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31
2
4434
1
3434
2
2
2
4234
1
3333
32
3222
23
3
4,3
23
233
23
23
1
3
23
23
23
321
hu
uhx
cuec
ji
uuh
xuhu
uuh
uuhx
uh
cuec
ji
uhu
xh
uhu
uhx
hu
n
n
n
n
n
Now with
ji
uhu
xx
uhu
huuhxx
uhu
hxuhx
uh
cucuec
ji
cuec
nn
nnnn
k
k
kjikijij
2
23
6
23
233
23
3621
1,2
32
1
32
222
32
41242123
1
2121
1
3
1
24433323
1
2323
22
1
22
3232
42243223
1
2122
3,2
23
6
323
2
23
232
1
uccuec
ji
uhu
xx
huxxuhuhux
xhuhx
uh
cucuec
nn
unn
n
nn
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32
uh
uhxx
huhuuhx
hu
uhxuhx
cucuec
ji
uuh
hxhuhxuxx
uuh
uhuuhxx
uuh
hxuxhuuhx
nn
nn
n
nnnn
nnnn
n
23
23
23
13
23
32
4,2
23
236
23
233
23
3632
2
2
22
2
44243423
1
2424
2
222
1
2
222
2
2
Now consider
uuh
ux
uuh
hxuxhux
uuh
uhhuhxuhxuxxx
cucucuec
ji
uhu
xhx
uhux
uhu
xx
uhu
xxx
cucucuec
ji
uhu
xhxhuhu
uhux
uhu
hxx
uhu
xxx
cucucuec
ji
cueC
nnnnnnnn
n
nn
n
nn
nnn
nn
nn
nnn
n
k
k
kjikijij
23
2
23
363
23
32636
3,1
23
23
23
2
23
6
23
6
2,1
23
2323
23
2
23
23
23
61
1
2
3
2
2
2
222222
431433132312
1
1313
32
32
32
3
32
2
1
2
32
1
421434132212
1
1212
32
323332
32
3
32
2
32
1
411431132112
1
1111
2
2
1
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33
4,1
23
23232
32223222
ji
uuh
xhuxhxuuxhxuhx nnnnnn
uh
xx
uh
x
uh
uhxx
uh
huxxx
cucucuec
unn
nnnnnn
23
2323
3
23
23
2
2
2
3
2
2
2
2
441434132412
1
1414
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.
uuh
hxxuuxxhuhxxu
uuh
xu
uuh
hxuxhu
uuh
xuhhuhxuhxuhxuxxx
uuh
xhuxhxuhxuhxuhx
xcxcxccx
uhu
xxhxx
uhu
x
uhu
xhx
uhu
xxhx
hu
xhx
xcxcxccx
uhu
uhuxxhxx
uhu
x
uhu
xhx
uhu
xhxx
uhu
xhxhuhu
xcxcxccx
nnn
nn
nnnnnn
nnnnnn
n
n
nnnnn
nn
nnnnn
23
3232
23
2
23
363
23
323636
23
232323
23
32
23
2
23
36
23
6623
23
2332
23
2
23
36
23
66
23
2323
2
2223
2
3
2
22
2
2222223
322222222
3
34
2
3323130
32
2
3
32
2
32
2
32
2
32
32
3
24
2
2322121
32
3223
32
3
32
2
32
2
32
323322
3
41
2
3121110
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34
uh
xxuhxx
uh
x
uh
xuhx
uh
xhuxx
uh
xhux
xcxcxccx
nn
nnnnn
23
2323
3
23
23
23
2
23
2
3
2
2
2
2
2
32
3
44
2
3424141
Hence forth,
11.2....2323
3232
23
32
23
2332
12
23
2
22223
.
32
23
32
3223
nnn
nnnn
nnn
nn
fuh
xxuhxxf
uuh
xxhuuxxhuxxu
unuhu
xxhxxy
uhu
uhuxxhxxxy y
Where (2.11) is the continuous form of one step Adams-Moulton scheme for k=1. If
(2.11) is evaluated at 1nx and a substitution 2
1u is made, the result is
hencefh
fh
yyxy nnnnn ,1144
22
1
12.2........................................................................................................4
2 11 21 nnnnn ff
hyyy
If on the other hand, equation (2.11) is differentiated with respect to x and then evaluated at
21
n
xx, the result obtained is
bfffh
yy
ffyh
yh
x
nnnn
n
nnnnny
12.2...........................................................................8524
8
1
8
533
21
21
21
1
2
1
,1
1
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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35
2
2
11
2
32
3210
3210
3210
1
unun
nn
nn
nnn
xx
xx
xx
xxx
D
The general forms of this system of matrix is given as :
13.2............................................................1100 unnnnn fxfxfxhyxxy
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:
1
1
1
0
12
2
12
2
2
2
0
13
1
13
13
1
;0
11
2
1
24
2
2
23
2
22
22
21
234
233
232
31
244
243
242
41
21
c
uuh
xxc
uh
xhuxc
uh
hxhuxxuhc
c
uuh
xc
uh
xuhc
uh
huhxc
c
uuhc
uuh
uc
uuh
uc
c
nn
nn
nnn
n
n
n
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36
16
23
6
2336
2
32
13
2
3222
12
xuh
xhxc
uh
xhxhuxhuxc
nn
nnnn
Following the same procedure as in the first case, the coefficients for the continuous schemes,
xandx jj are obtained and by similar substitution into equation (2.13), the desired continuous scheme
is obtained as follows;
14.2..............................16
32
16
32
16
1613121
2
23
12
223
2
2223
unnn
nnn
nnnn
n
fuuh
xxhxxf
uuh
xxhuxxu
fuuh
xxuuhxxuhxxuyxy
Putting u = ½ and proceeding to get our discrete forms as did earlier by evaluating (2.14) at these two
different points 1 nxx and 1 nn xx , we have:
a. at x = xn = xn+1, the result is
afff
hyy
hffh
fh
yy
nnnnn
nnnnn
15.2...........................................................................46
6
4
66
21
21
11
11
b. at x = x + xn+ ½ , the result is
bfff
hyy
nnnnn15.2......................................................................85
24 21
21 1
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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37
32
3
2
2
22
3
1
2
11
32
4
1
3
1
2
11
3
43210
43210
43210
43210
1
ununun
nnn
nnn
nnn
nnnn
xxx
xxx
xxx
xxx
xxxx
D
The general forms of D3 is given as
unnnnnn fxfxfxfxhyxxy
2211011
Appling the set of formulae (2.6)-(2.9) on this system gives the c-entries as follows
26
3
112
2412
24
12124
124
1
8
1
1,0
344
343
342
353
352
11314151
uh
huhxc
uh
huxc
uh
hxuhc
uuuhc
uhc
cccc
n
n
n
3
22233
22
3
22
35
3
22
34
3
22
33
3
222
32
3
1
45
2
2332
122
263
24
4323
12
4322
4
22622
12
uh
xhuxhuxhxxuhc
uuuh
hhxxc
uh
hxxhxxc
uh
hxxhuxuhc
uh
huxhhxuhuhc
uuuh
xc
nnnn
nn
nnnn
nnn
nn
n
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124
44
224
23644
112
8412853
24
3101212341824
1
23
1
22
3
44223
15
3
4442233
14
3
332244
13
3
44222343223
12
3
322
25
3
322
23
uuuh
hxhxhxc
uh
huhxhuxhuxhxc
hu
hxhuxhuxuxhhxc
uh
huxhhxhxxhuxhuxhuxc
uuh
xhxhxc
uh
xhxhuxhuxc
nnn
nnnn
nnnn
nnnnnn
nnn
nnnn
As in the previous cases, continuous schemes coefficients sxi '
and the sxi ' are obtained
and
substituted into the general form (2.16) to obtain the desired continuous schemes as follows;
17.2.............................................................................214
44
224
126143
112
8512243
24
10324121812431
3
42234
3
2
42234
3
1
42234
3
432234
1
uuuh
fhhxxhxxxx
uh
fhuxxuhxxuhxx
uh
fhuxxuhxxhuxx
uh
fhuxxuhhxxuhxxuxxyxy
unnnn
nnnn
nnnn
nnnnnn
Evaluating (2.17) at 2 nxx 2
3u , the resulting scheme is given as:
afffh
yynnnnn 18.2....................................................................................................4
6 232112
If (2.17) is evaluated at points unxx and 2
3u , we have
bffffh
yynnnnnn
18.2.............................................................................56546192 2
32
3 211
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39
However, if the point 2
3u is maintained and (2.17) is evaluated at x = xn , the result is another discrete
scheme of the form:
cffffh
yynnnnnn 18.2.........................................................................................472
6 23211
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:
3
2
2
22
3
1
2
11
32
432
4
1
3
1
2
!1
4
43210
43210
43210
1
1
nnn
nnn
nnn
unununun
nnnn
xxx
xxx
xxx
xxxx
xxxx
D
the general form of the matrix is given as:
19.2.......................................221011
nnnnunun fxfxfxhyxyxxy
As in the previous cases, matrix C is determined with the columns simplification yielding the continuous
schemes:
20.2..............1212
13112212
1213
222312122
12252121
2
121
1212
32
2
22
1
3
1
24
1
32
1
32
1
24
1
2
324
1422
1
22
1
4
1
422
42222
1
4
1
huu
fhxxuuhxxuuxxu
huuu
fhuuhxxuuuxxuu
fhuuxxuhuuu
yhxxxx
huuu
yhuuuhxxxxxy
nnnn
nnn
nnnnn
unnn
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40
Desired hybrid point of elevation remains 2
3U
and so (2.20) is evaluated to yield
afffh
yyy nnnnnn 21.2................................................................193212
1697 21122
3
at the point 2 nxx
If consideration at the point nxx
is made, we have:
bfffh
yyy nnnnnn 21.2........................................................................34894
1679 3112
3
Finding the differential expression with respect to x for (2.20) and evaluating the differential coefficient at
1 nxx , the result becomes:
cffffh
yynnnnnn 21.2..........................................................................56546
192 23
23 211
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
( a) The scheme (1.1) is said to be zero stable if no root of the polynomial
k
i
i
idp0
has modulo
greater than one and every root with modulo one must be distinct or simple.
( b) The order p and error constant 1pc for (1.1) could be defined thus:
If 0,0... 110 pp cccc ,then the principal local term error at knx is .11
1 n
pp
p xyhc
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41
1.3.........................21
!1
1
...21!
1
11
2
1
1
1
0
1
210
v
q
k
qqqq
qqqq
q
tvtktttq
tktttq
c
(c) A numerical method (1.1) is said to be consistent if ,1p 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:
affh
yyy nnnnn 12.2.........................................................................................4
2 11 21
From equation (2.12a)
4
1,
4
1,2,1,1 1010 2
1
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42
192
1
4
1.
6
12
6
11
24
1
!3
1
2
11
!4
1
08
12
8
11
6
1
!2
1
2
1
!3
1
04
12
2
11
2
1
2
1
!2
1
0114
1
2
121
2
1
0211
1
4
4
1
3
13
1
2
12
1011
100
21
21
21
21
21
c
c
c
c
c
Therefore the order p = 3 and
1.
01
021
21
21
21
2101
ei
p
Hence, the schemes is zero stable.
Example 3.1.2
Another scheme of k = 1 was given as:
2
12
1 8524
1
nnnnn fffh
yy
Here, 31
241
1245
002
12
1 ,,,,1,1 and and so
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43
384
1
144
1
144
1
384
1
2
1
!3
1
2
1!
4
1
048
1
48
1
24
1
2
1
48
1
12
1
24
1
2
1
48
1)
2
1(
2
1
2
1!
3
1
0)2
1(
3
1
24
1
8
1
2
1
2
1!
2
1
;024
12
2
1
2
1
;011
2
1
3
1
2
1
4
4
2
1
2
1
2
1
3
3
2
11
2
2
101
0
21
21
21
c
c
c
c
c
So that the order p = 3 and ;384
11 pc we seek the root of the characteristics polynomial as follows:
1
01 21
210
p
Hence, by definitions 3.1, we concluded that the method is zero stable.
We consider the scheme 2
32
3 56546192
211
nnnnnnffff
hyy
Here, 24
7,,
96
23,
192
5,
192
1,1,1
23
23 1201 and
Then observe that 0.................................................................................. 410 ccc
However,
3
4
23
2
4
1
5
23
15
1087.4
307224
3692
32
211
120
1
3072
4236
192
80
192
46
24
1
32
2431
120
1
2!4
1
!5
12
32
3
c
The order p=4 and .1087.4 3
1
pc the absolute root for the polynomial is thus computed;
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44
1,,,1,,0
01.
0
21
23
23
23 1
soandor
ei
p
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
Alagbe, S.A. (1999), “Derivation of a Class Multiplier Step Methods”, B.Sc. Project , Department of
Mathematics, University of Jos,
Awe, K.D. (1997), “Application of Adams Moulton Methods and its application in Block Forms”,
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)
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