^{1}

^{*}

^{1}

^{*}

The second derivative method which is A-stable is derived using Interpolation Collocation approach. The continuous method obtained are used to generate the main method and complementary methods to solve initial value problems of ordinary differential equation via boundary value technique. Numerical result obtained via the methods shows that the new method can compete with the existing ones in the literature.

Differential Equations (DEs) are important tools of solving real-life problems and a wide variety of natural phenomena are modeled into differential equations. DEs which normally arise in biological models, circuit theory models, circuit theory models, fluid and chemical kinetics models may or may not have exact solutions, thus a need for a numerical solution.

Consider the initial value problem of the form

y ′ = f ( x , y ) , y ( a ) = y 0 , x ∈ [ a , b ] (1)

Several numerical methods for the solution of (1) were proposed such as Tau method (Onumanyi [

In this paper, a continuous form of the second derivative multistep method shall be derived through a multistep collocation technique so that the method derived shall recover the Enright’s second derivative method and other possible method. The method will be in block and will be implemented on IVP. This paper will further discuss the implementation of the newly developed methods using Boundary Value Methods as it was also implemented in Axelsson and Verwer [

The second derivative multistep methods are derived using collocation technique as discussed in Ehigie [

The general second derivative formula for solving Equation (1) using k-step second derivative linear multistep method is of the form.

∑ j = 0 k α j y n + j = h ∑ j = 0 k β j f n + j + h 2 ∑ j = 0 k δ j g n + j (2)

where y n + j ≈ y ( x n + j h ) ,

f n + j ≡ f ( x n + j h , y ( x n + j h ) )

g n + j = d f ( x , y ( x ) ) d x | y = y n + j x = x n + j

x n is a discrete point at n, α j , β j and γ j are coefficients to be determined. To obtained the method of the form (1), y ( x ) is approximated by (2) where T n a chebyshev polynomial of the form

y ( x ) = ∑ n = 0 r a n T n ( x ) . (3)

Equation (3) will be considered for the derivation of the main and complementary methods for the two classes of continuous second derivatives multistep method of Enright which is a special case of (2).

Interpolating y ( x ) at point x n + k − 1 , collocating y ′ ( x ) at points x n + j , j = 0,1,2, ⋯ , k and collocating y ″ ( x ) at point x n + k , i.e.

x n + k − 1 = y n + k − 1

y ′ ( x ) = f n + j

y ″ ( x ) = g n + j , j = 0 , 1 , 2 , ⋯ , k

The system of equations generated are solved to obtained the coefficients of a j , j = 0,1,2, ⋯ , k + 4 which are used to generate the continous multistep method of Enright of the form

y ( x ) = y n + k − 1 + h ∑ j = 0 k β j f n + j + h 2 δ j g n + k (4)

Evaluating (4) at x = x n + k yields the second drivative multistep method of Enright also evaluating at x = x n + j , j = 0,1,2, ⋯ , k − 2 gives ( k − 1 ) methods, which will be called complementary methods to complete the k block for the system. The Enright’s method is obtained in the form.

y n + k = y n + k − 1 + h ∑ j = 0 k β j ( x ) f n + j + h 2 γ j g n + k (5)

The objective is to derive the multi-derivative main method of the form:

∑ j = 3 4 α j y n + j = h ∑ j = 0 4 β j f n + j + h 2 γ j g n + 4 (6)

where α j , β j , γ j are coefficients. In order to obtain Equation (6), we proceed by seeking an approximation to the exact solution by assuming a continuous solution y ( x ) of the form

y ( x ) = ∑ j = 0 6 a j T j ( x ) (7)

with first derivative given by

f ( x ) = y ′ ( x ) = ∑ j = 0 6 a j T ′ j ( x ) , (8)

with second derivative given by

g ( x ) = y ″ ( x ) = ∑ j = 0 6 a j T ″ j ( x ) , (9)

Interpolating (7) at x = x n + 3 , collocating (8) at x = x n + j , j = 0 ( 1 ) 6 and collocating (9) at x = x n + 4 where n is the grid index using maple we obtain the coefficients a ′ j s :

a 0 = y n + 3 − 22729 36864 h f n + 3 − 54275 589824 h f n − 103817 110592 h f n + 1 − 47411 49152 h f n + 2 + 198005 1769472 h f n + 4 − 6269 147456 h 2 g n + 4 ,

a 1 = 7459 49152 h f n + 1669 3072 h f n + 1 − 1489 4096 h f n + 2 + 899 3072 h f n + 3 − 6103 49152 h f n + 4 + 215 4096 h 2 g n + 4 ,

a 2 = 3365 73728 h f n + 1 − 731 24576 h f n + 3 + 15811 1179648 h f n + 4 + 955 32768 h f n + 2 − 571 98304 h 2 g n + 4 − 22981 393216 h f n ,

a 3 = 961 24576 h f n + 2 + 3181 294912 h f n − 571 18432 h f n + 3 − 589 18432 h f n + 1 + 3847 294912 h f n + 4 − 45 8192 h 2 g n + 4 ,

a 4 = 91 12288 h f n + 3 + 155 36864 h f n + 1 − 1967 589824 h f n + 4 − 119 16384 h f n + 2 + 71 49152 h 2 g n + 4 − 199 196608 h f n ,

a 5 = 133 491520 h f n + 4 + 19 40960 h f n + 2 − 1 8192 h 2 g n + 4 − 17 30720 h f n + 3 − 7 30720 h f n + 1 + 23 491520 h f n ,

a 6 = 1 294912 h 2 g n + 4 + 1 73728 h f n + 3 + 1 221184 h f n + 1 − 1 1179648 h f n − 1 98304 h f n + 2 − 25 3538944 h f n + 4

Substituting the values of a ′ j s in the approximate solution (7), will generate a continuous form of the scheme of the form:

y ( x ) = a 0 + a 1 ( 2 x − 1 ) + a 2 ( 8 x 2 − 8 x + 1 ) + a 3 ( 32 x 3 − 48 x 2 + 18 x − 1 ) + a 4 ( 128 x 4 − 256 x 3 + 160 x 2 − 32 x + 1 ) + a 5 ( 512 x 5 − 1280 x 4 + 1120 x 3 − 400 x 2 + 50 x − 1 ) + a 6 ( 2048 x 6 − 6144 x 5 + 6912 x 4 − 3584 x 3 + 840 x 2 − 72 x + 1 ) (10)

On evaluating (10) at x = x n + 4 , x = x n + 2 , x = x n + 1 , x = x n

y n + 4 − y n + 3 = h [ − 17 5760 f n + 1 45 f n + 1 − 41 480 f n + 2 + 47 90 f n + 3 + 3133 5760 f n + 4 ] − h 2 3 32 g n + 4 (11)

y n + 2 − y n + 3 = h [ − 11 1920 f n + 7 135 f n + 1 − 83 160 f n + 2 − 19 30 f n + 3 + 1831 17280 f n + 4 ] − h 2 11 288 g n + 4 (12)

y n + 1 − y n + 3 = h [ 1 90 f n − 17 45 f n + 1 − 19 15 f n + 2 − 17 45 f n + 3 + 1 90 f n + 4 ] (13)

y n − y n + 3 = h [ − 201 640 f n − 7 5 f n + 1 − 99 160 f n + 2 − 9 10 f n + 3 + 149 640 f n + 4 ] − h 2 3 32 g n + 4 (14)

Equations (11)-(14) will be taken as Boundary Value Method for k = 4 denoted as BVM4. To implement (10), we use a modified block method defined as follows:

h λ ∑ j = 1 q a i j y n + j λ = h λ ∑ j = 0 q e i j y n λ + h μ − λ [ ∑ j = 1 q d i j f n + ∑ j = 1 q b i j f n + j ] (15)

where λ is the power of the derivative of the continuous method and μ is the order of the problem to be solved; q = r + s . In vector notation, (15) can be written as

h λ a ¯ Y m = h λ e ¯ y m + h μ − λ [ d ¯ f ( y m ) + b ¯ F ( Y m ) ] (16)

The matrices a ¯ = ( a i j ) , b ¯ = ( b i j ) , e ¯ = ( e i j ) , d ¯ = ( d i j ) are constant coefficient matrices and Y m = ( y n + v i , y n + 1 , y ′ n + v i , y ′ n + 1 ) T , y m = ( y n − ( r − 1 ) , y n − ( r − 2 ) , ⋯ , y n ) T , F ¯ ( Y m ) = ( f n + v i , f n + j ) T and f ( y m ) = ( f n − i , ⋯ , f n ) , i = 1 , ⋯ , q . The normalized version of (16) is given by

A ¯ Y m = h λ E ¯ y m + h μ − λ [ D ¯ f ( y m ) + B ¯ F ( Y m ) ] (17)

The modified block formulae (15) and (16) are employed to simultaneously obtain value for y n + 1 , y n + 2 , y n + 3 , y n + 4 needed to implement (11). i.e. Combining (11)-(14) in the form of (15) and (16), yield the block method below:

( 1 − 1 0 0 0 − 1 1 0 0 − 1 0 1 0 − 1 0 0 ) ( y n + 4 y n + 3 y n + 2 y n + 1 ) = ( 0 0 0 1 ) ( y n ) + h ( 1 45 − 41 480 47 90 3133 5760 7 135 − 83 160 − 19 30 1831 17280 − 17 45 − 19 15 − 17 45 1 90 − 7 5 − 99 160 − 9 10 149 640 ) ( f n + 1 f n + 2 f n + 3 f n + 4 ) + h ( − 17 5760 − 11 1920 1 90 − 201 640 ) ( f n ) + h 2 ( − 3 32 − 11 288 0 − 3 32 ) ( g n + 4 )

using (16) to obtain the block solution as

( 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 ) = ( 1 1 1 1 ) ( y n ) + h ( 46 45 − 311 480 47 90 − 1277 5760 196 135 1 10 4 15 − 137 1080 7 5 99 160 9 10 − 149 640 64 45 8 15 64 45 14 45 ) ( f n + 1 f n + 2 f n + 3 f n + 4 ) + h ( 1873 5760 37 120 201 640 14 45 ) ( f n ) + h 2 ( 3 32 1 18 3 32 0 ) ( g n + 4 )

Written the above explicitly gives:

y n + 1 = y n + h [ 1873 5760 f n + 46 45 f n + 1 − 311 480 f n + 2 + 47 90 f n + 3 − 1277 5760 f n + 4 ] + h 2 3 32 g n + 4 (18)

y n + 2 = y n + h [ 37 120 f n + 196 135 f n + 1 + 1 10 f n + 2 + 4 15 f n + 3 − 137 1080 f n + 4 ] + h 2 1 18 g n + 4 (19)

y n + 3 = y n + h [ 201 640 f n + 7 5 f n + 1 + 99 160 f n + 2 + 9 10 f n + 3 − 149 640 f n + 4 ] + h 2 3 32 g n + 4 (20)

y n + 4 = y n + h [ 14 45 f n + 64 45 f n + 1 + 8 15 f n + 2 + 64 45 f n + 3 + 14 45 f n + 4 ] (21)

The objective is to derive the multi-derivative main method of the form:

∑ j = 4 5 α j y n + j = h ∑ j = 0 5 β j f n + j + h 2 γ j g n + 5 (22)

where α j , β j , γ j are coefficients are to be determined. In order to obtain Equation (22), I proceed by seeking an approximation to the exact solution by assuming a continuous solution y ( x ) of the form

y ( x ) = ∑ j = 0 7 a j T j ( x ) (23)

with first derivative given by

f ( x ) = y ′ ( x ) = ∑ j = 0 7 a j T ′ j ( x ) , (24)

with second derivative given by

g ( x ) = y ″ ( x ) = ∑ j = 0 7 a j T ″ j ( x ) , (25)

Interpolating (23) at x = x n + 4 , collocating (24) at x = x n + j , j = 0 ( 1 ) 5 and collocating (25) at x = x n + 5 where n is the grid index using maple we obtain the coefficients a j s:

a 0 = y n + 4 − 463807 368640 h f n + 3 − 316403 3686400 h f n − 181051 184320 h f n + 1 − 903161 1105920 h f n + 2 − 265139 737280 h f n + 4 + 13987 5529600 h f n + 5 + 623 184320 h 2 g n + 5 ,

a 1 = 237263 1638400 h f n + 153505 262144 h f n + 1 − 138739 294912 h f n + 2 + 27969 65536 h f n + 3 − 20081 65536 h f n + 4 + 7059707 58982400 h f n + 5 − 47011 983040 h 2 g n + 5 ,

a 2 = 327 8192 h f n + 1 − 47111 819200 h f n + 717 16384 h f n + 2 − 799 16384 h f n + 3 + 1237 32768 h f + n + 4 − 6207 409600 h f n + 5 + 251 40960 h 2 g n + 5 ,

a 3 = 56431 4915200 h f n − 9485 262144 h f n + 1 + 44029 884736 h f n + 2 − 8695 196608 h f n + 3 + 2053 65536 h f n + 4 − 2152541 176947200 h f n + 5 + 14293 2949120 h 2 g n + 5 ,

a 4 = − 509 409600 h f n + 23 4096 h f n + 1 − 89 8192 h f n + 2 + 99 8192 h f n + 3 − 153 16384 h f n + 4 + 767 204800 h f n + 5 − 31 20480 h 2 g n + 5 ,

a 5 = 961 983040 h f n + 4 − 1561 3932160 h f n + 1 − 1603 3932160 h f n + 5 + 443 491520 h f n + 2 + 11 65536 h 2 g n + 5 − 1129 983040 h f n + 3 + 73 983040 h f n ,

a 6 = 71 3686400 h f n + 5 − 5 147456 h f n + 2 − 1 122880 h 2 g n + 5 + 7 147456 h f n + 3 + 1 73728 h f n + 1 − 17 7372800 h f n − 13 294912 h f n + 4 ,

a 7 = 1 6881280 h 2 g n + 5 − 1 1376256 h f n + 3 − 1 5505024 h f n + 1 + 1 34406400 h f n + 1 2064384 h f n + 2 + 1 1376256 h f n + 4 − 137 412876800 h f n + 5

Substituting the values of a ′ j s in the approximate solution (23), I obtain a continuous form of the scheme of the form:

y ( x ) = a 0 + a 1 ( 2 x − 1 ) + a 2 ( 8 x 2 − 8 x + 1 ) + a 3 ( 32 x 3 − 48 x 2 + 18 x − 1 ) + a 4 ( 128 x 4 − 256 x 3 + 160 x 2 − 32 x + 1 ) + a 5 ( 512 x 5 − 1280 x 4 + 1120 x 3 − 400 x 2 + 50 x − 1 ) + a 6 ( 2048 x 6 − 6144 x 5 + 6912 x 4 − 3584 x 3 + 840 x 2 − 72 x + 1 ) + a 7 ( 8192 x 7 − 28672 x 6 + 39424 x 5 − 26880 x 4 + 9408 x 3 − 1568 x 2 + 98 x − 1 ) (26)

On evaluating (26) at x = x n + 5 , x = x n + 3 , x = x n + 2 , x = x n + 1 , x = x n we generate the main methods and complementary methods which invariably complete the boundary value techniques.

y n + 5 − y n + 4 = h [ 41 25200 f n − 529 40320 f n + 1 + 373 7560 f n + 2 − 1271 10080 f n + 3 + 2837 5040 f n + 4 + 317731 604800 f n + 5 ] − h 2 863 10080 g n + 5 (27)

y n + 3 − y n + 4 = h [ 19 8400 f n − 89 4480 f n + 1 + 677 7560 f n + 2 − 1933 3360 f n + 3 − 323 560 f n + 4 + 48467 604800 f n + 5 ] − h 2 271 10080 g n + 5 (28)

y n + 2 − y n + 4 = h [ − 1 630 f n + 53 2520 f n + 1 − 382 945 f n + 2 − 773 630 f n + 3 − 263 630 f n + 4 + 221 7560 f n + 5 ] − 1 126 h 2 g n + 5 (29)

y n + 1 − y n + 4 = h [ 33 2800 h f n − 1737 4480 f n + 1 − 337 280 f n + 2 − 1023 1120 f n + 3 − 339 560 f n + 4 + 2201 22400 f n + 5 ] − 39 1120 h 2 g n + 5 (30)

y n − y n + 4 = h [ − 158 525 f n − 52 35 f n + 1 − 344 945 f n + 2 − 176 105 f n + 3 − 2 35 f n + 4 − 548 4725 f n + 5 ] + 16 315 h 2 g n + 5 (31)

Equations (27)-(31) will be taken as Boundary Value method for k = 5 and denoted as BVM5. The modified block formulae (15) and (16) are employed to simultaneously obtain value for y n + 1 , y n + 2 , y n + 3 , y n + 4 , y n + 5 needed to implement (27), i.e., combining (27)-(31) in the form of (15) and (16), yield the block method below:

( 1 − 1 0 0 0 0 − 1 1 0 0 0 − 1 0 1 0 0 − 1 0 0 1 0 − 1 0 0 0 ) ( y n + 5 y n + 4 y n + 3 y n + 2 y n + 1 ) = ( 0 0 0 0 1 ) ( y n ) + h ( − 529 40320 373 7560 − 1271 10080 2837 5040 317731 604800 − 89 4480 677 7560 − 1933 3360 − 323 560 48467 604800 − 53 2520 − 382 945 − 773 630 − 263 630 221 7560 − 1737 4480 − 337 280 − 1023 1120 − 339 560 2201 22400 − 52 35 − 344 945 − 176 105 − 2 35 − 548 4725 ) ( f n + 5 f n + 4 f n + 3 f n + 2 f n + 1 )

+ h ( 41 25200 19 8400 − 1 630 33 2800 − 158 525 ) ( f n ) + h 2 ( − 863 10080 − 271 10080 − 1 126 − 39 1120 16 315 ) ( g n + 5 )

using (16) to obtain the block solution 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 ) = ( 1 1 1 1 1 ) ( y n ) + h ( 4919 4480 − 6347 7520 2563 3360 − 307 560 129571 604800 3797 2520 − 38 945 283 630 − 227 630 5489 37800 6567 4480 127 280 1233 1120 − 291 560 4393 22400 52 35 344 945 − 176 105 2 35 548 4725 11875 8064 625 1512 3125 2016 625 1008 15515 24192 ) ( f n + 1 f n + 2 f n + 3 f n + 4 f n + 5 ) + h ( 2627 8400 943 3150 849 2800 158 525 305 1008 ) ( f n ) + h 2 ( − 863 10080 − 37 630 − 87 1120 − 16 315 275 2016 ) ( g n + 5 )

Written the above explicitly gives:

y n + 1 = y n + h [ 2627 8400 f n + 4919 4480 f n + 1 − 6347 7560 f n + 2 + 2563 3360 f n + 3 − 307 560 f n + 4 + 129571 604800 f n + 5 ] − 863 10080 h 2 g n + 5 (32)

y n + 2 = y n + h [ 943 3150 f n + 3797 2520 f n + 1 − 38 945 f n + 2 + 283 630 f n + 3 − 227 630 f n + 4 + 5489 37800 f n + 5 ] − 37 630 h 2 g n + 5 (33)

y n + 3 = y n + h [ 849 2800 f n + 6567 4480 f n + 1 + 127 280 f n + 2 + 1233 1120 f n + 3 − 291 560 f n + 4 + 4393 22400 f n + 5 ] − 87 1120 h 2 g n + 5 (34)

y n + 4 = y n + h [ 158 525 f n + 52 35 f n + 1 + 344 945 f n + 2 + 176 105 f n + 3 + 2 35 f n + 4 + 548 4725 f n + 5 ] − 16 315 h 2 g n + 5 (35)

y n + 5 = y n + h [ 305 1008 f n + 11875 8064 f n + 1 + 625 1512 f n + 2 + 3125 2016 f n + 3 + 625 1008 f n + 4 + 15515 24192 f n + 5 ] − 275 2016 h 2 g n + 5 (36)

Basic properties of the main schemes and their associated block schemes, are analysed to establish their validity. These properties, namely: order, error constant, consistency and zero stability reveal the nature of convergence of the schemes. The region of absolute stability of the methods have also been obtained in this section.

Order of the Scheme

Let the linear difference operator L associated with the continuous multi-derivative method (2) be defined as

L [ y ( x ) ; h ] = ∑ j = 0 k { α j y ( x n + j h ) − h β j y ′ ( x n + j h ) − h 2 β j y ″ ( x n + j h ) ; j = 1 , 2 , ⋯ , m } (37)

where y ( x ) is an arbitrary test function that is continuously differentiable in the interval [ a , b ] . Expanding y ( x n + j h ) and y ″ ( x n + j h ) , j = 0,1,2, ⋯ in Taylor series about x n and collecting like terms in h and y gives:

L [ y ( x ) ; h ] = C 0 y ( x ) + C 1 h y ( 1 ) ( x ) + C 2 h 2 y ( 2 ) ( x ) + ⋯ + C p h p y ( p ) ( x ) + ⋯ (38)

Definition (4.1.1). The difference operator L and the associated continuous multistep method (2) are said to be of order p C 0 = C 1 = C 2 = C 3 = ⋯ = C p = C p = 0 , C p + 1 ≠ 0 Lambert [

Order of the Block Scheme

The order of the block will be defined following the method of Chollom [

Let the multi-derivative method be defined by:

∑ j α i j ( μ ) ( t ) y n + j = h ∑ j β i j ( μ ) ( t ) f n + j + h 2 ∑ j γ i j ( μ ) ( t ) g n + j , (39)

where μ is the degree of the derivative of the continuous cofficients α i j ( t ) , γ i j and β i j ( t ) .

The linear difference operator, L, associated with the implicit block hybrid one step method (15) is described by the formula:

L [ y ( x ) ; h ] = ∑ j [ α ¯ j y ( x n + j h ) − h β ¯ j y ′ ( x n + j h ) − h 2 γ j ( x n + j h ) ] (40)

where y ( x ) is an arbitrary test function continuously differentiable on [ a , b ] . Expanding y ( x n + j h ) and y ″ ( x n + j h ) , in Taylor series and collecting terms in (41) gives:

L [ y ( x ) ; h ] = C ¯ 0 y ( x ) + C ¯ 1 h y ( 1 ) ( x ) + C ¯ 2 h 2 y ( 2 ) ( x ) + ⋯ + C ¯ p h p y ( p ) ( x ) (41)

where the C ¯ i , i = 0 , 1 , ⋯ , p are vectors.

Consistency

Given a continuous multi-derivative method (2), the first and second characteristic polynomials are defined as

ρ ( z ) = ∑ j = 0 k α j z j (42)

σ ( z ) = ∑ j = 0 k β j z j (43)

where z is the principal root, α k ≠ 0 and α 0 2 + β 0 2 ≠ 0 .

Definition (4.1.2). The continuous multi-derivative method (2) is said to be consistent if it satisfies the following conditions:

1) the order p ≥ 1 ;

2) ∑ j = 0 k α j = 0 ;

3) ρ ( 1 ) = ρ ′ ( 1 ) = 0 ;

4) ρ ″ ( 1 ) = 2 ! σ ( 1 ) .

Definition (4.1.3). According to Gurjinder [

Applying the definitions for order of the method on the main scheme (11) for four-step method and order of the block methods on (18)-(21) likewise order of the main scheme (27) for five-step method and order of the block methods on (32)-(36) it was derived that the main scheme (11) is of order p = 6 with Error Constant C p + 1 = 41 30240 and the additional schemes are of uniform order p = 6 , likewise for the five-step method the main scheme (27) is of order p = 7 with Error Constant C p + 1 = 731 846720 . And the additional schemes is of uniform order p = 7 . The order of four-step and five-step block methods are in tabular form as indicated in

The main schemes and additional schemes for four-step method and five-step method are consistent since they have order p ≥ 1 .

Zero Stability and Convergent

It is known from literature that the stability of linear multi-step method determine the manner in which the error is propagated as the numerical computation

Method | Order | Error Constant C p + 1 |
---|---|---|

(18) | 6 | − 8 945 |

(19) | 6 | − 11 1120 |

(20) | 6 | − 1 126 |

(21) | 6 | − 3327 30240 |

Method | Order | Error Constant C p + 1 |
---|---|---|

(32) | 7 | 1375 169344 |

(33) | 7 | 16 2205 |

(34) | 7 | 257 31360 |

(35) | 7 | 187 26460 |

(36) | 7 | 2633 282240 |

proceeds hence the investigation of the zero-stability property is necessary.

Definition (4.1.4). The continuous multi-derivative method (2) is said to be zero-stable if no root of the first characteristic polynomial ρ ( z ) has modulus greater than one, and if every root of modulus one has multiplicity not greater than one.

Definition (4.1.5). The implicit hybrid block method (17) is said to be zero stable if the roots z s , s = 1, ⋯ , n of the first characteristic polynomial ρ ¯ ( z ) , defined by

ρ ¯ ( z ) = det [ z A ¯ − E ¯ ] (44)

satisfies | z s | ≤ 1 and every root with | z s | = 1 has multiplicity not exceeding two in the limit as h → 0 . Lambert [

In order to study the efficiency of the developed schemes, we present some numerical experiments with the following five examples. The Four-Step Methods (4step) and Five-Step Methods (5step) were applied to solve the following test problems:

Example 1. The linear problem by Enright in Ehigie [

y 1 = − 0.1 y 1 , y 1 ( 0 ) = 1

y 2 = − 10 y 2 , y 2 ( 0 ) = 1

y 3 = − 100 y 3 , y 3 ( 0 ) = 1

y 4 = − 1000 y 4 , y 4 ( 0 ) = 1

is solved in the range 0 ≤ x ≤ 10 .

Exact solution: y 1 = exp ( − 0.10 x ) , y 2 = e x p ( − 10 x ) , y 3 = e x p ( − 100 x ) , y 4 = e x p ( − 1000 x ) .

Example 2.

Considering a real life problem, which is an epidemical model (SIR model) that computes the theoretical numbers of people infected with a contagious illness in a closed population over time. The name of this class of models derive from the fact that they involves couples equations relating the numbers of susceptiple people S(t), number of people infected I(t), number of people who have recovered R(t). This is a good and simple model for many infectious desease including measles, mumps and rubella Sunday [

d s d t = μ ( I − S ) − β I S (45)

d I d t = − μ I − γ I + β I S (46)

d R d t = − μ R + γ I (47)

where μ , γ are positive parameters. Defined y to be

y = S + I + R

Adding Equations (45) to (47) to obtain the following evolution equation for y

y ′ = μ ( 1 − y ) (48)

taking μ = 0.5 and attaching initial condition y ( 0 ) = 0.5 (for a particular closed population) to obtain,

y ′ ( t ) = 0.5 ( 1 − y ) , y ( 0 ) = 0.5 (49)

whose exact solution is y ( t ) = 1 − 0.5 e − 0.5 t .

Example 3.

The non-linear system solved by Wu and Xia cited in Ehigie [

y ′ = − 1002 y 1 + 1000 y 2 2 , y 1 ( 0 ) = 1 ,

y ′ 2 = y 1 − y 2 ( 1 + y 2 ) , y 2 ( 0 ) = 1

Exact solution y 1 ( x ) = e − 2 x , y 2 ( x ) = e − x

Example 4.

Considering the moderately stiff problem solve by Jia-Xiang and Jiao-Xun cited in Ehigie [

y ′ = − y − 10 z , y ( 0 ) = 1 ,

z ′ = − 10 y − z , z ( 0 ) = 0 ,

Exact solution y ( x ) = e − x cos 10 x , z ( x ) = e − x sin 10 x

In this work, new multi-derivative multistep methods of order (6, 7) are proposed for the solution of first order Initial Value Problems. The main method and additional methods for cases (k = 4, 5) are obtained from the same continuous scheme derived via interpolation and collocation procedures. The stability properties and region of the method were also discussed. The methods were then applied in block form as simultaneous numerical integrators over non-overlapping interval. Numerical results obtained using the proposed block form show that it is attractive for the solution of linear and non-linear problems.

In Tables 3-7, numerical examples were used to implement the new derived methods, comparing the accuracy of method with those of Ehigie, Okunuga Jator and Sofoluwe [

x | Step | y 1 | y 2 | y 3 | y 4 |
---|---|---|---|---|---|

Enright [ | 1000 | 4.1 × 10 − 7 | - | - | - |

Ehigie [ | 1000 | 4.2 × 10 − 16 | 2.1 × 10 − 10 | 0 | 0 |

Four-Step | 1000 | 0 | 0 | 0 | 0 |

Five-Step | 1000 | 0 | 0 | 0 | 0 |

x | Four-Step | Five-Step | Sunday [ |
---|---|---|---|

0.1 | 3.766 × 10 − 12 | 1.54 × 10 − 13 | 2.0 × 10 − 11 |

0.2 | 2.498 × 10 − 12 | 1.08 × 10 − 13 | 3.0 × 10 − 11 |

0.3 | 3.013 × 10 − 12 | 1.23 × 10 − 13 | 1.0 × 10 − 10 |

0.4 | 2.408 × 10 − 12 | 1.00 × 10 − 13 | 2.0 × 10 − 10 |

0.5 | 5.374 × 10 − 12 | 1.09 × 10 − 13 | 1.0 × 10 − 10 |

0.6 | 4.225 × 10 − 12 | 2.24 × 10 − 13 | 2.0 × 10 − 10 |

0.7 | 4.538 × 10 − 12 | 1.82 × 10 − 13 | 1.0 × 10 − 10 |

0.8 | 3.943 × 10 − 12 | 1.90 × 10 − 13 | 2.0 × 10 − 10 |

0.9 | 6.274 × 10 − 12 | 1.67 × 10 − 13 | 3.0 × 10 − 10 |

1.0 | 5.242 × 10 − 12 | 1.70 × 10 − 13 | 3.0 × 10 − 10 |

No. of Steps | Four-Step max ‖ y i − y ( x i ) ‖ | Four-Step max ‖ z i − z ( x i ) ‖ | Ehigie [ | Ehigie [ |
---|---|---|---|---|

125 | 1.121 × 10 − 11 | 7.224 × 10 − 11 | 8.33 × 10 − 06 | 1.32 × 10 − 6 |

250 | 1.225 × 10 − 13 | 3.551 × 10 − 13 | 1.13 × 10 − 07 | 1.36 × 10 − 08 |

500 | 1.00 × 10 − 15 | 6.780 × 10 − 15 | 8.19 × 10 − 12 | 6.30 × 10 − 12 |

No. of Steps | Five-Step max ‖ y i − y ( x i ) ‖ | Five-Step max ‖ z i − z ( x i ) ‖ Math_274# | Ehigie [ | Ehigie [ |
---|---|---|---|---|

125 | 1.635 × 10 − 12 | 4.795 × 10 − 12 | 8.33 × 10 − 06 | 1.32 × 10 − 6 |

250 | 2.0 × 10 − 15 | 1.6 × 10 − 14 | 1.13 × 10 − 07 | 1.36 × 10 − 08 |

500 | 0 | 0 | 8.19 × 10 − 12 | 6.30 × 10 − 12 |

Method | Number of Steps | max ‖ y i − y ( x i ) ‖ | max ‖ z i − z ( x i ) ‖ |
---|---|---|---|

Wu-Xia [ | 500 | 2.56 × 10 − 07 | 8.02 × 10 − 08 |

Four-Step | 500 | 5.68 × 10 − 09 | 5.38 × 10 − 09 |

Five-Step | 500 | 4.22 × 10 − 13 | 9.89 × 10 − 13 |

The study was not funded by any grant, it was funded by the authors.

The authors declare that there is no conflict of interest.

Areo, E.A. and Edwin, O.A. (2020) Multi-Derivative Multistep Method for Initial Value Problems Using Boundary Value Technique. Open Access Library Journal, 7: e6063. https://doi.org/10.4236/oalib.1106063