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									Mathematical Theory and Modeling                                                                  www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.10, 2012



     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 ni  h r f nv . …………………………………………………(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 ( xnv , ynv ) .

 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 n1  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 n1  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 n1  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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Mathematical Theory and Modeling                                                                                                                www.iiste.org
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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 m1,1 
                      12             t 1, 2               h 02         h m1, 2 
 C
        02                                                                                 ..........................2.5
                                                                                   
                                                                                          
    0 j  m
                    1 j m            t 1, j  m          h 0 j  m     h m1,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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Mathematical Theory and Modeling                                                                                                       www.iiste.org
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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 
                                                          
     
    yx    0 x  y n   u x y nu  h 0 x  f n  1 x  f n1 

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
            j2
                    d12
            u12         xn
                    l11
            j3
                    d13
            u13         xn
                           2

                    l11
            j4
                    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
               j2
                      e12
              e12 
               1
                          0
                      l11
               j3
                      e13
              e13 
               1
                          0
                      l11
               j4
                      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 j4

          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 j4
                                              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              u h 3  2u  
                          3 x n  uh 2 2           
                    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      23 x  uh 
                           2 3n
                     u h22
                             u h 3  2u 
                      3h  2 x n 
                    2 3
                     u h 3  2u 
                   i j3
                   c33  e33  u 34c 42 
                          1


                       1                 2  u  
                          3 x n  uh  2            
                       uh                 uh 3  2u  

                   
                       u       
                            3 h  3u  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


                                                  
                                                             
                       1 2 x n  uh 6 x n  3h  3 x n  3 x n uh  2u 2 h 2
                          
                                                         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 
                    i2 j

                   c 22  e1  u 23c32  u 24c 42
                           21

                        1               3h  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 nu
                           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  2u h 2 
                                           3 x n uh  u 2 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   32 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



              
                          2
                               
                 x n 3x n  2ux n h
                                     
                                       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



          
              2x  x             n   3  3hx  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
                                                                                                                
                                                  uh3  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,

                      2x  x n 3  3h x  x n 2  u 2 h 3 3  2u 
                                                                                2x  x n 3  3hx  x n 2  .
                                                                                                                
              yx                                                      yn                                   yn  u
                      
                                       u h 3  2u 
                                         2 3
                                                                         
                                                                               
                                                                                        u h 3  2u 
                                                                                          2 3
                                                                                                                 
                                                                                                                 
                                    3         2
                                                                   
                 2  u x  x n   3  u hx  x n   2u  3u h x  x n 
                
                                                            2        2        2
                                                                                               x  x n   uhx  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
                   
                   yxn 1    y n  2 y n 12 
                                                                h      h
                                                                  f n  f n1, hence
                                                                4      4

              y n1  y n  2 y n 12 
                                                   h
                                                      f n1  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 :


yx    0 x y n  h 0 x  f n  1 x  f n1   n x  f nu ............................................................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;



                                                          
                   2u  1x  x n 3  3h u 2  1  x  x n 2  6uh 2 u  1x  x n 
                  
y x   1y n  
                                                                                                
                                                                                                 fn
                  
                                                  6uh u  1
                                                       2
                                                                                                
                                                                                                
     2u x  x n   3u hx  x n  
                                                    2 x  x n   3hx  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  xn1        and
                                             xn  xn1 , 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 n1  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 xn2         3xn 2    4 xn2 
                0 1                                   2
                                                             4 x n u 
                                                                 3
                                   2 x n u       3x n u            

   The general forms of D3 is given as

   
   yx   1 x  y n1  h 0 x  f n  1 x  f n1   2 x  f n 2   n x  f nu 

       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  42  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     
                              
                    3x n h  2 x n h 2  x n
                        2                   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
                                                                                                       
                                  12u  1h 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   1y n 1 
                                   3x  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


    3x  x    n
                         4
                              42  u h x  x n   12uh 2 x  x n   5  8u h 4 f n 1
                                                    3                   2
                                                                                                           
                                               12h 3 u  1


   3x  x            n
                             4
                                   4hu  1 x  x n   6uh 2  x  x n   2u  1h 4 f n  2
                                                          3                                2
                                                                                                            
                                                  24h 3 u  2


  x  x   n      4x  x n  h  4x  x n  h 2  h 4 f n u
                     4                     3                      2
                                                                              
                                                                  .............................................................................2.17 
                          4uh 3 u  1u  2
Evaluating     (2.17) at x  x n  2 u  3 2 the resulting scheme is given as:
                                            ,

y n 2  y n1 
                             h
                             6
                                                             
                               f n1  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 n1 
                              h
                             192
                                                                                          
                                  f n  46 f n1  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 n1  y n 
                h
                6
                                                          
                  2 f n  7 f n1  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 xn2 
                                                                   3
                                       2 xn2       3xn 2             



the general form of the matrix is given as:


yx   1 x y n1   u x y nu  h 0 x  f n   n x  f n1   2 x  f n 2 .......................................2.19

As in the previous cases, matrix C is determined with the columns simplification yielding the continuous
schemes:


yx  
         x  x      4  2x  xn1 2 h 2  u  12 u 2  2u  1h 4 y nu
                      n 1

                               u  12 u 2  2u  1h 4

  x  xn1 4  2x  xn1 2 h 2 yn1   2u  5x  x 4  2u 2  2u  1h 3 f
         u  12 u 2  2u  1h 4                              n 1                      n





  u 2  2u  2x  xn1 4  u  1u 2  2u  1x  xn1 3h  2u 2  2u  2h 3 f n1
                                     3u  1u 2  2u  1h 3


   2u  1x  xn1 4  2u 2  2u  1x  xn1 3 h  u  13u  1x  xn1 2 h 2 f n2                       ..............2.20 
                                                                    
                                                   12 u 2  2u  1 h 3




                                                                    39
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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 n1  16 y n 3 2 
                                      h
                                         f n  32 f n1  19 f n2 ................................................................2.21a 
                                     12
at the point x  x
                   n 2


If consideration at the point x  x is made, we have:
                                   n


9 y n1  7 y n  16 y n 3 2 
                                   h
                                      9 f n  48 f n1  3 f n3 ........................................................................2.21b
                                   4


Finding the differential expression with respect to x for (2.20) and evaluating the differential coefficient at
xx       the result becomes:
       n 1   ,

y n1  y n 3 2 
                      h
                     192
                                                                    
                         f n  46 f n1  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




                                                                     40
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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 n1  y n  2 y n 12 
                             h
                                f n1  f n .........................................................................................2.12a 
                             4



From equation (2.12a)

                                          1        1
 0  1, 1  1,   2,  0 
                     1                       , 1 
                         2
                                          4         4




                                                                      41
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           c0   0   1   12  1  1  2  0
                            
           c1    1   12    0   1 
                        1
                       2    
                    1  1
            1  2    1  1  0
                        
                    2  4
              1                  
                             2

           c2    1   1   1   1
                          2
              2! 
                       2       
            1 1           1
            1   2    0
            2 2           4
                1                  
                               3

           c3      1   1   1   1 1
                            2
                3! 
                         2        2!
             1 1            1
            1   2    0
             6 8            8
                 1  1         1
                            4

           c 4  1     1 2    1
                 4!   2 
                    
                                 3!
                                
               1  1          1 1
                1   2   .
              24  6          6 4
                 1
           
                192

Therefore the order p = 3 and

                                        
                       p    1   0   12           1
                                                               2




                        1  2 2  0
                                         1




                            1
                                   2
                                         
                                       1  0
                      i.e   1

Hence, the schemes is zero stable.

Example 3.1.2

Another scheme of k = 1 was given as:


y n  y n  12 
                   h
                   24
                       
                      f n 1  5 f n  8 f n 12   
Here,  0  1,  12  1,  0  5 24 , 1             1
                                                           24      , and ,  12  13 and so



                                                                             42
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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 n1  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;




                                                                  43
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             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

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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