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In this paper, block procedure for some k-step linear multi-step methods, using the Legendre polynomials as the basis functions, is proposed. Discrete methods were given which were used in block and implemented for solving the initial value problems, being continuous interpolant derived and collocated at grid points. Some numerical examples of ordinary differential equations were solved using the derived metho ds to show their validity and the accuracy. The numerical results obtained show that the proposed method can also be efficient in solving such problems.

Many problems in celestial and quantum mechanics, nuclear, theoretical physics, astrophysics, quantum chemistry and molecular dynamics, are of great interest to scientists and engineers. These problems are mathematically modelled by using ordinary differential equation of the form:

f ( x , y , y ′ , y ″ , ⋯ , y ( n ) ) , y ( a ) = y 0 , y ′ ( a ) = y 1 , ⋯ , y ( n − 1 ) ( a ) = y n (1)

where on the interval [ a , b ] has given rise to two major discrete variable methods namely, one step and multistep methods commonly known as linear multi-step methods. Many authors have worked on the direct solution of (1), among which are Lambert [

In Awoyemi [

Furthermore, many researchers had developed interest on improving the numerical solution of initial value problems of ordinary differential equation. Consequently, the development of a class of methods called block method is one of the outcomes. This was proposed by Milne [

In this section, we consider the approximate solution of the form

y k ( x ) = ∑ i = 0 k c i ψ i ( x ) .

Perturbing the equation above, we have

∑ i = 0 k c i ψ i ( x ) = f ( x , y ) + λ L k ( x ) (2)

where, ψ i ( x ) = x i , i = 0 , 1 , ⋯ , k and L k ( x ) is the Legendre polynomial of degree k, which is defined on the interval [ − 1,1 ] , and can be determined with the aid of the recurrence formula:

L i + 1 ( x ) = 2 i + 1 i + 1 x L i ( x ) − i i + 1 L i − 1 ( x ) , i = 1 , 2 , ⋯ (3)

So that

L 0 ( x ) = 1 , L 1 ( x ) = x , L 2 ( x ) = 3 x 2 − 1 2 , L 3 ( x ) = 5 x 3 − 3 x 2 , L 4 ( x ) = 35 x 4 − 30 x 2 + 3 2

We define a shifted Legendre polynomials by introducing the change of variable

x = 2 x ¯ − ( x n + k + x n ) x n + k − x n , k = 1 , 2 , 3 , 4 (4)

For k = 1

using Equation (4), taking L 1 ( x ) = x , and collocating at x n and x n + 1 , we obtain

L 1 ( x n ) = − 1 , L 1 ( x n + 1 ) = 1 (5)

hence,

L 1 ( x n ) = 2 x n − x n + 1 − x n x n + 1 − x n = − 1 (6)

and,

L 1 ( x n + 1 ) = 2 x n + 1 − x n + 1 − x n x n + 1 − x n = 1 (7)

Deducing ψ 0 ( x ) = 0 , ψ 1 ( x ) = 1 from Equation (1), it follows that Equation (2) becomes

f ( x , y ) = c 1 − λ L 1 ( x ) (8)

Solving the above systems we obtain

λ = 1 2 ( f n − f n + 1 ) , c 1 = f n − λ , c 0 = y n − x n ( f n − λ )

The required numerical scheme of the method will be obtained if we collocate y ( x ) = c 0 + c 1 x at x = x n + 1 and substitute c 0 , c 1 , λ as follows

y n + 1 = y n + h 2 ( f n + 2 + f n ) (9)

k = 2

taking L 2 ( x ) = 1 2 ( 3 x 2 − 1 ) , and collocating at x n , x n + 1 , and x n + 2

we get

L 2 ( x n ) = 1 , L 2 ( x n + 1 ) = − 1 2 , L 2 ( x n + 2 ) = 1

From Equation (1), it can be deduced that ψ 0 ( x ) = 0 , ψ 1 ( x ) = 1 , ψ 2 ( x ) = 2 x , then Equation (2) becomes

f ( x , y ) = c 1 + 2 x c 2 − λ L 2 ( x ) (10)

Collocating Equation (10) at x n + i , i = 0 , 1 , 2 and interpolate

y k ( x ) = ∑ i = 0 k c i ψ i ( x ) , x n ≤ x ≤ x n + k (11)

at x = x n , we get a system of four equations with c i ( i = 0 , 1 , 2 ) and parameter λ

y n = c 0 + c 1 x n + c 2 x n 2 f n = c 1 + 2 c 2 x n − λ f n + 1 = c 1 + 2 c 2 x n + 1 − 1 2 λ f n = c 1 + 2 c 2 x n + 2 − λ (12)

Hence, solving Equation (12), we get

λ = 1 3 ( − f n + 2 f n + 1 − f n + 2 ) c 0 = − 1 12 h ( − 12 h y n + 8 t n h f n + 1 − 4 t n h f n + 2 + 8 t n h f n − 3 t n 2 f n + 2 + 3 t n 2 f n ) c 1 = 1 6 h ( 4 h f n + 1 − 2 h f n + 2 + 4 h f n − 3 t n f n + 2 + 3 t n f n ) c 2 = − 1 4 h ( f n − f n + 2 ) (13)

From y k ( x ) = ∑ i = 0 k c i ψ i ( x ) , we have

y ¯ = c 0 + c 1 x + c 2 x 2 (14)

Hence, the required numerical scheme is obtained by collocating Equation (14) above at x = x n + 1 and substituting c 0 , c 1 , c 2 , λ as follows

y n + 1 = y n + h 12 ( 5 f n + 8 f n + 1 − f n + 2 ) (15)

k = 3

Taking the polynomial L 3 = 1 2 ( 5 x 3 − 3 x ) and use the Equation (4), then

collocating this at x n , x n + 1 , x n + 2 and x n + 3 , we obtain

L 3 ( x n ) = − 1 , L 3 ( x n + 1 ) = 11 27 , L 3 ( x n + 2 ) = − 11 27 , L 3 ( x n + 3 ) = 1 . From Equation

(1), it can be deduced that ψ 0 ( x ) = 0 , ψ 1 ( x ) = 1 , ψ 2 ( x ) = 2 x , ψ 3 ( x ) = 3 x 2 , then Equation (2) is reduced to the form

f ( x , y ) = c 1 + 2 x c 2 + 3 c 3 x 2 − λ L 3 ( x ) (16)

Hence, collocating Equation (16) at x n + i , i = 0 , 1 , 2 , 3 and interpolate Equation (11) at x = x n , we get the system of equations with c i , ( i = 0 , 1 , 2 , 3 ) and parameter λ

y n = c 0 + c 1 x n + c 2 x n 2 + c 3 x n 3 f n = c 1 + 2 c 2 x n + 3 c 3 x n 2 + λ f n + 1 = c 1 + 2 c 2 x n + 2 + 3 c 3 x n + 1 2 − 11 27 λ f n + 2 = c 1 + 2 c 2 x n + 2 + 3 c 3 x n + 2 2 + 11 27 λ f n + 3 = c 1 + 2 c 2 x n + 3 + 3 c 3 x n + 3 2 − λ (17)

Solving the above system of equations, we obtain

λ = 9 40 ( f n − 3 f n + 1 + 3 f n + 2 − f n + 3 )

c 0 = − 1 120 ( − 81 t n h 2 f n + 2 + 27 t n f n + 3 h 2 + 56 t n 2 h 2 f n + 81 t n h 2 f n + 1 + 93 t n h 2 f n − 18 t n 2 h f n + 1 + 34 t n 2 h f n + 3 − 72 t n 2 h f n + 2 − 120 t n 2 h f n + 2 − 10 t n 3 f n + 2 + 10 t n 3 h f n + 3 − 10 t n 3 f n + 1 )

c 1 = 1 120 ( 68 t n h f n + 3 + 112 t n f n h − 36 t n 2 h f n + 1 − 144 t n h f n + 2 + 81 h 2 f n + 1 + 27 h 2 f n + 3 + 93 h 2 h f n − 81 h 2 f n + 2 − 30 t n 2 f n + 2 + 30 t n 2 f n + 3 + 30 t n 2 f n − 30 t n 2 f n + 1 )

c 2 = − 1 60 h 2 ( − 15 t n f n + 2 + 15 t n f n − 15 t n f n + 1 − 36 h f n + 2 + 28 h f n + 17 h f n + 3 − 9 h f n + 1 )

c 3 = − 1 12 h 2 ( f n − f n + 1 − f n + 2 + f n + 3 )

From y k ( x ) = ∑ i = 0 k c i ψ i ( x ) , we have

y ¯ = c 0 + c 1 x + c 2 x 2 + c 3 x 3 (18)

Hence, the required numerical scheme is obtained by collocating Equation (18) above at x = x n + 1 and substituting c 0 , c 1 , c 2 , c 3 , λ as follows

y n + 1 = y n + h 120 ( 47 f n + 89 f n + 1 − 19 f n + 2 + 3 f n + 3 ) (19)

k = 4

In this case, we take the polynomial L 4 = 1 8 ( 35 x 4 − 30 x 2 + 3 ) and use the

Equation (4), then collocating this at x n , x n + 1 , x n + 2 , x n + 3 and x n + 4 , we obtain

L 4 ( x n ) = 1 , L 4 ( x n + 1 ) = − 37 128 , L 4 ( x n + 3 ) = 3 8 , L 4 ( x n + 4 ) = 1 . From Equation (1),

we can deduce that ψ 0 ( x ) = 0 , ψ 1 ( x ) = 1 , ψ 2 ( x ) = 2 x , ψ 3 ( x ) = 3 x 2 , ψ 4 ( x ) = 4 x 3 , then Equation (2) is reduced to the form

f ( x , y ) = c 1 + 2 x c 2 + 3 c 3 x 2 + 4 c 4 x 3 − λ L 4 ( x ) (20)

Hence, collocating Equation (20) at x n + i , i = 0 , 1 , 2 , 3 , 4 and interpolate Equation (11) at x = x n , we get the system of equations with c i , ( i = 0 , 1 , 2 , 3 , 4 ) and parameter λ

y n = c 0 + c 1 x n + c 2 x n 2 + c 3 x n 3 + c 4 x n 4 f n = c 1 + 2 c 2 x n + 3 c 3 x n 2 + 4 c 4 x n 3 − λ f n + 1 = c 1 + 2 c 2 x n + 1 + 3 c 3 x n + 1 2 + 4 c 4 x n + 1 3 + 37 128 λ f n + 2 = c 1 + 2 c 2 x n + 2 + 3 c 3 x n + 2 2 + 4 c 4 x n + 2 3 − 3 8 λ f n + 3 = c 1 + 2 c 2 x n + 3 + 3 c 3 x n + 3 2 + 4 c 4 x n + 3 3 + 37 128 λ f n + 4 = c 1 + 2 c 2 x n + 4 + 3 c 3 x n + 4 2 + 4 c 4 x n + 3 3 − λ (21)

Solving the above system of equations with a suitable method, we obtain

λ = 16 105 ( − f n + 4 f n + 1 − 6 f n + 2 + 4 f n + 3 − f n + 4 )

c 0 = − 1 5040 h 3 ( 768 x n h 3 f n + 4 − 4272 x n f n h 3 − 3072 x n h 3 f n + 3 − 3072 x n h 3 f n + 1 + 4608 x n h 3 f n + 2 + 2400 x n 2 h 2 f n + 1 − 3330 x n 2 h 2 f n + 1290 x n 2 h 2 f n + 4 − 4320 x n 2 h 2 f n + 3 + 1520 x n 3 h f n + 1 − 1840 x n 3 h f n + 3 + 670 x n 3 h f n + 4 − 1010 x n 3 h f n + 660 x n 3 h f n + 2 + 3960 x n 2 h 2 f n + 2 + 5040 y n h 3 + 210 x n 4 f n + 1 − 210 x n 4 f n + 3 + 105 x n 4 f n + 4 − 105 x n 4 f n )

c 1 = 1 840 h 3 ( − 1320 x n h 2 f n + 2 − 330 x n 2 f n + 2 h + 505 x n 2 f n h − 335 x n 2 f n + 4 h + 920 x n 2 f n + 3 h + 760 x n 2 h f n + 1 + 1440 x n h 2 f n + 3 − 430 x n h 2 f n + 4 + 1110 x n h 2 f n + 800 h 2 x n f n + 1 − 128 h 3 h f n + 4 − 768 h 3 h f n + 2 + 512 h 3 f n + 1 + 512 h 3 f n + 3 + 712 h 3 f n − 140 x n 3 f n + 1 + 140 x n 3 f n + 3 − 70 x n 3 f n + 4 + 70 x n 3 f n )

c 2 = − 1 168 h 3 ( − 66 x n h f n + 2 + 101 x n h f n − 67 x n h f n + 4 + 184 x n h f n + 3 − 152 x n h f n + 1 + 43 h 2 f n + 4 + 111 f n h 2 − 42 x n 2 f n + 1 + 42 x n 2 f n + 1 + 42 x n 2 f n + 3 − 21 x n 2 f n + 4 + 21 x n 2 f n − 80 h 2 f n + 1 + 144 h 2 f n + 3 − 132 h 2 f n + 3 )

c 3 = − 1 504 h 3 ( − 84 x n f n + 1 + 84 x n f n + 3 − 42 x n f n + 4 + 42 x n f n − 152 h f n + 1 + 184 h f n + 3 − 67 h f n + 4 + 101 h f n − 66 h f n + 2 )

c 4 = − 1 48 h 3 ( f n − 2 f n + 1 + 2 f n + 3 − f n + 4 )

From y k ( x ) = ∑ i = 0 k c i ψ i ( x ) , we have

y ¯ = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 (22)

Hence, the required numerical scheme is obtained by collocating Equation (22) above at x = x n + 1 and substituting c 0 , c 1 , c 2 , c 3 , c 4 , λ as follows

y n + 1 = y n + h 5040 ( 1847 f n + 4162 f n + 1 − 1308 f n + 2 + 382 f n + 3 − 43 f n + 4 ) (23)

Formulating the Block Scheme of Cases k = 2, 3 and 4

If k = 2

We collocate Equation (14) at x = x n + 1 , x n + 2 , x n + 3 to give us

y n + 1 = y n + h 12 ( 5 f n + 8 f n + 1 − f n + 2 ) y n + 2 = y n + h 3 ( f n + 4 f n + 1 + f n + 2 ) y n + 3 = y n + h 4 ( 8 f n + 1 − f n + f n + 2 ) (24)

If k = 3

We collocate Equation (18) at x = x n + 1 , x n + 2 , x n + 3 , x n + 4 to give us

y n + 1 = y n + h 120 ( 47 f n + 89 f n + 1 − 19 f n + 2 + 3 f n + 3 ) y n + 2 = y n + h 60 ( 21 f n + 77 f n + 1 + 23 f n + 2 − f n + 3 ) y n + 3 = y n + h 8 ( 3 f n + 9 f n + 1 + 9 f n + 2 + 3 f n + 3 ) y n + 4 = y n + h 30 ( 29 f n − 7 f n + 1 + 47 f n + 2 + 51 f n + 3 ) (25)

If k = 4

We collocate Equation (22) at x = x n + 1 , x n + 2 , x n + 3 , x n + 4 , x n + 5 to give us

y n + 1 = y n + h 5040 ( 1847 f n + 4162 f n + 1 − 1308 f n + 2 + 382 f n + 3 − 43 f n + 4 ) y n + 2 = y n + h 90 ( 29 f n + 124 f n + 1 + 24 f n + 2 + 4 f n + 3 − f n + 4 ) y n + 3 = y n + h 560 ( 179 f n + 377 f n + 1 + 444 f n + 2 + 334 f n + 3 − 31 f n + 4 ) y n + 4 = y n + h 45 ( 14 f n + 64 f n + 1 + 24 f n + 2 + 64 f n + 3 + 14 f n + 4 ) y n + 5 = y n + h 1008 ( − 253 f n + 3322 f n + 1 − 1308 f n + 2 + 1222 f n + 3 + 2057 f n + 4 ) (26)

The schemes developed belong to the class of Linear Multi-step Method (LMM) of the form

∑ j = 0 k α j ( x ) y ( x n + j ) = h ∑ j = 0 k β j ( x ) f ( x n + j ) (27)

Equation (27) is a method associated with a linear difference operator

L [ y ( x ) ; h ] = ∑ j = 0 k ( α j y ( x + j h ) = h β j y ′ ( x + j h ) ) (28)

where y ( x ) is continuously differentiable on the interval [ a , b ] , and the Taylor series expansion about the point x is expressed as

L [ y ( x ) ; h ] = c 0 y ( x ) + c 1 h y ′ ( x ) + c 2 h 2 y ″ ( x ) + ⋯ + c q h q y q ( x ) (29)

In line with [

when k = 2 , P = 3 and C p + 1 = 0.041667

when k = 3 , P = 3 and C p + 1 = 0.016

when k = 4 , P = 2 and C p + 1 = 0.041667

The scheme can be expressed as:

[ 1 0 0 0 1 0 0 0 1 ] [ y n + 1 y n + 2 y n + 3 ] = [ 0 0 1 0 0 1 0 0 1 ] [ y n − 2 y n − 1 y n ] + h [ 8 12 − 1 12 0 4 3 1 3 0 2 5 4 0 ] [ f n + 1 f n + 2 f n + 3 ] + [ 0 0 5 12 0 0 1 3 0 0 − 1 4 ] [ f n − 2 f n − 1 f n ]

where,

A ( 0 ) = [ 1 0 0 0 1 0 0 0 1 ] , A ( 1 ) = [ 0 0 1 0 0 1 0 0 1 ] , B ( 0 ) = [ 8 12 − 1 12 0 4 3 1 3 0 2 5 4 0 ]

and

B ( 1 ) = [ 0 0 5 12 0 0 1 3 0 0 − 1 4 ]

The first characteristics polynomial of the scheme is

ρ ( λ ) = d e t [ λ A 0 − A 1 ]

ρ ( λ ) = d e t [ ( λ 0 0 0 λ 0 0 0 λ ) − ( 0 0 1 0 0 1 0 0 1 ) ] = d e t [ λ 0 − 1 0 λ − 1 0 0 λ − 1 ]

| λ 0 − 1 0 λ − 1 0 0 λ − 1 | = 0

λ 2 ( λ − 1 ) = 0

λ 1 = λ 2 or λ 3 = 1

A block method is said to be stable as h → 0 if the roots of the first characteristics polynomial defined by

ρ λ = d e t [ λ A 0 − A 1 ]

satisfies | r s | = 1

The scheme can be expressed as

[ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ] [ y n + 1 y n + 2 y n + 3 y n + 4 ] = [ 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 ] [ y n − 3 y n − 2 y n − 1 y n ] + h [ 89 120 − 19 120 1 40 0 124 90 4 15 2 45 − 1 90 77 60 23 60 − 1 60 0 9 8 9 8 3 8 0 − 7 30 47 30 17 10 0 ] [ f n + 1 f n + 2 f n + 3 f n + 4 ] + [ 0 0 0 47 120 0 0 0 7 20 0 0 0 3 8 0 0 0 29 30 ] [ f n − 3 f n − 2 f n − 1 f n ]

where,

A ( 0 ) = [ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ] , A ( 1 ) = [ 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 ] , B ( 0 ) = [ 89 120 − 19 120 1 40 0 77 60 23 60 − 1 60 0 9 8 9 8 3 8 0 − 7 30 47 30 17 10 0 ]

and

B ( 1 ) = [ 0 0 0 47 120 0 0 0 7 20 0 0 0 3 8 0 0 0 29 30 ]

The first characteristics polynomial of the scheme is

ρ ( λ ) = d e t [ λ A 0 − A 1 ]

ρ ( λ ) = d e t [ ( λ 0 0 0 0 λ 0 0 0 0 λ 0 0 0 0 λ ) − ( 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 ) ] = d e t [ λ 0 0 − 1 0 λ 0 − 1 0 0 λ − 1 0 0 0 0 λ − 1 ]

| λ 0 0 − 1 0 λ 0 − 1 0 0 λ − 1 0 0 0 λ − 1 | = 0

λ 3 ( λ − 1 ) = 0

λ 1 = λ 2 = λ 3 = 0 or λ 4 = 1

A block method is said to be stable as h → 0 if the roots of the first characteristics polynomial defined by

ρ λ = d e t [ λ A 0 − A 1 ]

satisfies | r s | = 1

The scheme can be expressed as

[ 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 ] [ y n + 1 y n + 2 y n + 3 y n + 4 y n + 5 ] = [ 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ] [ y n − 4 y n − 3 y n − 2 y n − 1 y n ] + h [ 56150 144 − 84084 144 56042 144 − 14009 144 0 124 90 4 15 2 45 − 1 90 0 754 560 444 560 334 560 − 31 560 0 64 45 8 9 64 45 14 45 0 3322 1008 − 1308 1008 1222 1008 2057 1008 0 ] [ f n + 1 f n + 2 f n + 3 f n + 4 f n + 5 ]

+ [ 0 0 0 0 − 13955 144 0 0 0 0 29 90 0 0 0 0 179 560 0 0 0 0 14 45 0 0 0 0 − 253 1008 ] [ f n − 4 f n − 3 f n − 2 f n − 1 f n ]

where,

A ( 0 ) = [ 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 ] , A ( 1 ) = [ 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ] , B ( 0 ) = [ 56150 144 − 84084 144 56042 144 − 14009 144 0 124 90 4 15 2 45 − 1 90 0 754 560 444 560 334 560 − 31 560 0 64 45 8 9 64 45 14 45 0 3322 1008 − 1308 1008 1222 1008 2057 1008 0 ]

and

B ( 1 ) = [ 0 0 0 0 − 13955 144 0 0 0 0 29 90 0 0 0 0 179 560 0 0 0 0 14 45 0 0 0 0 − 253 1008 ]

The first characteristics polynomial of the scheme is

ρ ( λ ) = d e t [ λ A 0 − A 1 ]

ρ ( λ ) = d e t [ ( λ 0 0 0 0 λ 0 0 0 0 λ 0 0 0 0 λ ) − ( 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 ) ] = d e t [ λ 0 0 − 1 0 λ 0 − 1 0 0 λ − 1 0 0 0 0 λ − 1 ]

| λ 0 0 0 − 1 0 λ 0 0 − 1 0 0 λ 0 − 1 0 0 0 λ − 1 0 0 0 0 λ − 1 | = 0

λ 4 ( λ − 1 ) = 0

λ 1 = λ 2 = λ 3 = λ 4 = 0 or λ 5 = 1

In order to confirm the accuracy and efficiency of the scheme, we consider the following initial value problems: Tables 1-4.

N | x | YC | YEX | |YEX − YC| |
---|---|---|---|---|

0 | 0.000000 | 1.000000 | 1.000000 | 0.000000 |

1 | 0.100000 | 0.904874 | 0.904837 | 3.7 × 10^{−5} |

2 | 0.200000 | 0.818800 | 0.818730 | 7.0 × 10^{−5} |

3 | 0.300000 | 0.740889 | 0.740818 | 7.1 × 10^{−5} |

4 | 0.400000 | 0.670418 | 0.670320 | 9.8 × 10^{−5} |

5 | 0.500000 | 0.606653 | 0.606531 | 1.22 × 10^{−4} |

6 | 0.600000 | 0.548956 | 0.548811 | 1.45 × 10^{−4} |

7 | 0.700000 | 0.496749 | 0.496585 | 1.64 × 10^{−4} |

8 | 0.800000 | 0.449511 | 0.449328 | 1.83 × 10^{−4} |

9 | 0.900000 | 0.406768 | 0.406569 | 1.99 × 10^{−4} |

10 | 1.000000 | 0.368093 | 0.367879 | 2.14 × 10^{−4} |

N | x | YC | YEX | |YEX − YC| |
---|---|---|---|---|

0 | 0.000000 | 1.000000 | 1.000000 | 0.000000 |

1 | 0.100000 | 0.905695 | 0.904837 | 8.58 × 10^{−4} |

2 | 0.200000 | 0.820365 | 0.818730 | 1.635 × 10^{−3} |

3 | 0.300000 | 0.743155 | 0.740818 | 2.337 × 10^{−3} |

4 | 0.400000 | 0.673292 | 0.670320 | 2.972 × 10^{−3} |

5 | 0.500000 | 0.610078 | 0.606531 | 3.547 × 10^{−3} |

6 | 0.600000 | 0.552879 | 0.548811 | 4.068 × 10^{−3} |

7 | 0.700000 | 0.501124 | 0.496585 | 4.539 × 10^{−3} |

8 | 0.800000 | 0.463649 | 0.449328 | 1.4321 × 10^{−2} |

9 | 0.900000 | 0.429739 | 0.406569 | 2.3170 × 10^{−2} |

10 | 1.000000 | 0.399057 | 0.367879 | 3.1178 × 10^{−2} |

N | x | YC | YEX | |YEX − YC| |
---|---|---|---|---|

0 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

1 | 0.100000 | 0.004993 | 0.004988 | 5.0 × 10^{−6} |

2 | 0.200000 | 0.019806 | 0.019801 | 5.0 × 10^{−6} |

3 | 0.300000 | 0.044003 | 0.044003 | 0.00000 |

4 | 0.400000 | 0.076890 | 0.076884 | 6.0 × 10^{−6} |

5 | 0.500000 | 0.117509 | 0.117503 | 6.0 × 10^{−6} |

6 | 0.600000 | 0.164732 | 0.164729 | 3.0 × 10^{−6} |

7 | 0.700000 | 0.217303 | 0.217295 | 8.0 × 10^{−6} |

8 | 0.800000 | 0.273858 | 0.273851 | 7.0 × 10^{−6} |

9 | 0.900000 | 0.333026 | 0.333023 | 3.0 × 10^{−6} |

10 | 1.000000 | 0.393477 | 0.393469 | 8.0 × 10^{−6} |

N | x | YC | YEX | |YEX − YC| |
---|---|---|---|---|

0 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

1 | 0.100000 | 0.004434 | 0.004988 | 5.5 × 10^{−4} |

2 | 0.200000 | 0.019809 | 0.019801 | 8.0 × 10^{−6} |

3 | 0.300000 | 0.044009 | 0.044003 | 6.0 × 10^{−6} |

4 | 0.400000 | 0.076891 | 0.076884 | 7.0 × 10^{−6} |

5 | 0.500000 | 0.129887 | 0.117503 | 1.2 × 10^{−2} |

6 | 0.600000 | 0.168705 | 0.164729 | 3.9 × 10^{−3} |

7 | 0.700000 | 0.216646 | 0.217295 | 6.5 × 10^{−4} |

8 | 0.800000 | 0.272938 | 0.273851 | 9.1 × 10^{−4} |

9 | 0.900000 | 0.344677 | 0.333023 | 1.1 × 10^{−2} |

10 | 1.000000 | 0.395906 | 0.393469 | 2.4 × 10^{−3} |

Problem 1:

y ′ = − y ( x ) , h = 0.1 , y ( 0 ) = 1 (30)

Exact solution: y ( x ) = e − x (see K.M. Abualnaja, 2015).

YC: approximate solution

YEX: exact solution

Problem 2:

y ′ ( x ) = − x ( 1 − y ) , h = 0.1 , y ( 0 ) = 0 (31)

Exact solution: y ( x ) = 1 − e − x 2 2 (see K.M. Abualnaja, 2015).

YC: approximate solution

YEX: exact solution

In this research work, a class of implicit block collocation methods for the direct solution of initial value problems of general first order ordinary differential equations was developed using Legendre collocation approach. The collocation technique yielded a consistent and zero stable implicit block multi-step method with continuous coefficients. The method is implemented without the need for the development of correctors.

Okedayo, T.G., Owolanke, A.O., Amumeji, O.T. and Adesuyi, M.P. (2018) Modified Legendre Collocation Block Method for Solving Initial Value Problems of First Order Ordinary Differential Equations. Open Access Library Journal, 5: e4565. https://doi.org/10.4236/oalib.1104565