This paper presents a study on the development and implementation of a second derivative method for the solution of stiff first order initial value problems of ordinary differential equations using method of interpolation and collocation of polynomial approximate solution. The results of this paper bring some useful information. The constructed methods are A-stable up to order 8 . As it is shown in the numerical examples, the new methods are superior for stiff systems.
We considered development of second derivative method for the solution of
y ′ = f ( x , y ) , y ( x n ) = y 0 , x n ≤ x ≤ x N (1)
where x n is the initial points, y : [ x n , x N ] → R m , f : [ x n , x N ] × R → R m is continuous and at least twice differentiable. We seek the solution on equidistant set of points defined on the integration interval x n = x 0 < x 1 < ⋯ < x N = b ,
x n = x 0 + n h , n = 0,1,2, ⋯ , N − 1 , h = b − 1 N − 1 , N is a positive integer.
A potentially good numerical method for the solution of stiff systems must have good accuracy and reasonably wide region of absolute stability. A-Stability requirement is the minimum criteria on the choice of suitable methods. The search for higher order A-stable linear multistep method is carried out in two ways; firstly, the use of higher derivatives of the approximate solution and secondly, the inclusion of additional stages of off grid points Ezzeddine and Hojjati [
Several authors such as Enright [
The aim of this paper is to develop a class of second derivative linear multistep method with varying step-lengths which are A-stable with large region of absolute stability (see Figures 1-3). The three methods recovered are tested on some numerical examples and their results compared with each other in order to determine how to fix the varying step-lengths to obtain the best results as shown Tables 1-4.
We considered the approximate solution of the form
y ( x ) = ∑ n = 0 j a n x n (2)
where a n ’s are constants to be determined. The ith derivative of (2) gives
y ( i ) ( x ) = ∑ n − i j n ( n − 1 ) ( n − 2 ) ⋯ ( n − i ) a n x n − i (3)
Imposing the following conditions on (2)
y ( x n ) = y n , y ′ ( x n + j ) = f n + j , j = 0 , 1 , 2 , ⋯ , s , y ″ ( x n + j ) = g n + j , j = 0 , 1 , 2 , ⋯ , τ
evaluating (2) at x n , first derivative of (3) at x n + j , j = 0 , 1 , ⋯ , s and second derivative of (3) at x n + j , j = 0 , 1 , ⋯ , τ give a system of equations in the form
A X = U (4)
where
X = [ 1 x n x n 2 x n 3 ⋯ x n j 0 1 2 x n 3 x n 2 ⋯ j x n j − 1 0 1 2 x n + 1 3 x n + 1 2 ⋯ j x n + 1 j − 1 ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ 0 1 2 x n + s 3 x n + s 2 ⋯ j x n + s j − 1 0 0 2 6 x n ⋯ j ( j − 1 ) x n j − 2 0 0 2 6 x n + 1 ⋯ j ( j − 1 ) x n + 1 j − 2 ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 2 6 x n + τ ⋯ j ( j − 1 ) x n + τ j − 2 ]
A = [ a 0 a 1 ⋯ a s + τ + 2 ] T , U = [ y n f n ⋯ f n + s g n ⋯ g n + τ ] T
| x | h | yi | Case I | Case II | Case III |
|---|---|---|---|---|---|
| 10 | 0.1 | y1 | 4.28E−22 | 3.94E−19 | 2.13E−18 |
| y2 | 4.49E−24 | 4.15E−21 | 2.24E−20 | ||
| 0.05 | y1 | 1.65E−24 | 1.67E−21 | 9.23E−21 | |
| y2 | 1.29E−26 | 1.76E−23 | 9.72E−23 | ||
| 0.001 | y1 | 4.96E−24 | 4.96E−24 | 4.14E−24 | |
| y2 | 5.17E−26 | 5.17E−26 | 4.52E−26 | ||
| 15 | 0.1 | y1 | 2.89E−26 | 2.68E−23 | 1.45E−22 |
| y2 | 3.03E−28 | 2.83E−25 | 1.53E−24 | ||
| 0.05 | y1 | 2.27E−28 | 1.14E−25 | 6.29E−25 | |
| y2 | 2.37E−30 | 1.20E−27 | 6.62E−27 | ||
| 0.001 | y1 | 4.04E−28 | 4.80E−28 | 3.53E−28 | |
| y2 | 4.34E−30 | 5.13E−30 | 3.94E−30 |
| x | h | yi | Case I | Case II | Case III |
|---|---|---|---|---|---|
| 10 | 0.1 | y1 | 2.44E−19 | 4.49E−17 | 2.51E−11 |
| y2 | 1.54E−09 | 2.23E−07 | 5.48E−08 | ||
| 0.05 | y1 | 4.07E−19 | 3.79E−11 | 5.42E−19 | |
| y2 | 3.91E−10 | 1.19E−07 | 2.89E−08 | ||
| 0.001 | y1 | 8.13E−20 | 5.42E−19 | 1.89744−19 | |
| y2 | 1.58E−11 | 2.50E−08 | 6.04E−09 | ||
| 15 | 0.1 | y1 | 2.12E−21 | 4.54E−19 | 2.54E−18 |
| y2 | 1.05E−13 | 1.52E−11 | 3.73E−12 | ||
| 0.05 | y1 | 4.24E−21 | 3.81E−21 | 5.51E−21 | |
| y2 | 2.66E−14 | 8.10E−12 | 1.97E−12 | ||
| 0.001 | y1 | 6.35E−22 | 4.87E−21 | 1.06E−21 | |
| y2 | 1.08E−15 | 1.70E−12 | 4.11E−13 |
Solving (4) using cramer’s method for the unknown constants, substitute it into (2), we obtain continuous linear multistep method in the form
y n + t = y n + h ∑ j = 0 s β j ( t ) f n + j + h 2 ∑ j = 0 τ η j ( t ) g n + j (5)
subject to h ∑ j = 0 s β j ( t ) = t h , β j ( t ) and η j ( t ) are polynomial of degree s + τ + 1 . Evaluating (5) at selected grid points give a block method in the form
ζ ( 1 ) Y m + 1 = ζ ( 0 ) Y m + h ( η ( 0 ) F m + η ( 1 ) F m + 1 ) + h 2 ( γ ( 0 ) G m + γ ( 1 ) G m + 1 ) (6)
where
| x | h | yi | Case I | Case II | Case III |
|---|---|---|---|---|---|
| 10 | 0.1 | y1 | 1.37E−11 | 2.50E−11 | 1.40E−11 |
| y2 | 4.69E−10 | 1.78E−07 | 4.79E−08 | ||
| 0.05 | y1 | 7.11E−12 | 1.61E−11 | 8.86E−12 | |
| y2 | 2.67E−10 | 1.15E−07 | 3.10E−08 | ||
| 0.001 | y1 | 1.06E−13 | 7.01E−13 | 3.40E−13 | |
| y2 | 5.71E−12 | 4.77E−09 | 1.28E−09 | ||
| 15 | 0.1 | y1 | 6.20E−16 | 1.13E−15 | 6.34E−16 |
| y2 | 3.16E−12 | 1.19E−09 | 3.23E−10 | ||
| 0.05 | y1 | 3.23E−16 | 7.30E−16 | 4.02E−16 | |
| y2 | 1.80E−12 | 7.77E−10 | 2.09E−10 | ||
| 0.001 | y1 | 4.79E−18 | 3.18E−17 | 1.54E−17 | |
| y2 | 3.85E−14 | 3.21E−11 | 8.64E−12 |
| x | h | yi | Case I | Case II | Case III |
|---|---|---|---|---|---|
| 15 | 0.1 | y1 | 1.60E−09 | 2.27E−10 | 2.23E−10 |
| y2 | 6.94E−18 | 6.94E−18 | 1.04E−17 | ||
| y3 | 8.88E−16 | 8.88E−16 | 1.78E−15 | ||
| y4 | 1.11E−16 | 1.11E−16 | 1.67E−16 | ||
| 0.05 | y1 | 1.51E−09 | 2.28E−10 | 2.23E−10 | |
| y2 | 1.39E−17 | 1.04E−17 | 1.04E−17 | ||
| 8.88E−16 | 8.88E−16 | 8.88E−16 | |||
| 2.78E−17 | 8.33E−17 | 8.33E−17 | |||
| 0.001 | y1 | 1.43E−09 | 2.48E−10 | 2.23E−10 | |
| y2 | 6.94E−18 | 3.47E−17 | 2.43E−17 | ||
| 8.88E−16 | 3.55E−15 | 2.22E−15 | |||
| 2.78E−17 | 1.94E−16 | 8.33E−17 |
Y m + 1 = [ y n + 1 y n + 2 ⋯ y n + s ] T , F m = [ f n − 1 f n − 2 ⋯ f n ] T
F m + 1 = [ f n + 1 f n + 2 ⋯ f n + s ] T , Y m = [ y n − 1 y n − 2 ⋯ y n ] T
G m = [ g n − 1 g n − 2 ⋯ g n ] T , G m + 1 = [ g n + 1 g n + 2 ⋯ g n + τ ] T
Writing (6) in the form
F ( Y m + 1 ) = Y m + 1 − ζ ( 0 ) Y m − h ( η ( 0 ) F m + η ( 1 ) F m + 1 ) − h 2 ( γ ( 0 ) G m + γ ( 1 ) G m + 1 ) = 0 (7)
Solving the system of nonlinear equations using Newton Raphson’s method gives
Y m + 1 κ + 1 = Y m + 1 κ − ( J κ ) − 1 F ( Y m + 1 κ ) (8)
where J κ is the Jacobian matrix. The necessary and sufficient condition for convergence of (8) is that the spectral radius of the inverse of the Jacobian matrix | ρ ( J κ − 1 ) | < 1
J = | ∂ F ( y n + 1 ) ∂ y n + 1 ∂ F ( y n + 2 ) ∂ y n + 1 ⋯ ∂ F ( y n + s ) ∂ y n + 1 ∂ G ( y n + 1 ) ∂ y n + 1 ∂ G ( y n + 1 ) ∂ y n + 1 ⋯ ∂ G ( y n + τ ) ∂ y n + 1 ∂ F ( y n + 1 ) ∂ y n + 2 ∂ F ( y n + 2 ) ∂ y n + 2 ⋯ ∂ F ( y n + s ) ∂ y n + 2 ∂ G ( y n + 1 ) ∂ y n + 2 ∂ G ( y n + 2 ) ∂ y n + 2 ⋯ ∂ G ( y n + τ ) ∂ y n + 2 ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋱ ⋮ ∂ F ( y n + s ) ∂ y n + s ∂ F ( y n + s ) ∂ y n + s ⋯ ∂ F ( y n + s ) ∂ y n + s ∂ G ( y 2 ) ∂ y n + s ∂ G ( y n + 2 ) ∂ y n + s ⋯ ∂ G ( y n + τ ) ∂ y n + s |
Specification of the MethodIn this paper, we consider grid points x n + j , j = 0 < u < v < w , hence the (6) reduces to
ζ ( 1 ) = [ 1 0 0 0 1 0 0 0 1 ] , ζ ( 0 ) = [ 0 0 1 0 0 1 0 0 1 ] , η ( 0 ) = [ 0 0 θ 11 0 0 θ 21 0 0 θ 31 ]
η ( 1 ) = [ θ 12 θ 13 θ 14 θ 22 θ 23 θ 24 θ 32 θ 33 θ 34 ] , γ ( 0 ) = [ 0 0 θ 15 0 0 θ 25 0 0 θ 35 ] , γ ( 1 ) = [ θ 16 θ 17 θ 18 θ 26 θ 27 θ 28 θ 36 θ 37 θ 30 ]
Y m + 1 = [ y n + u y n + v y n + w ] T , Y m = [ y n − 1 y n − 2 y n ] T ,
F m = [ f n − 1 f n − 2 f n ] T , F m + 1 = [ f n + u f n + v f n + w ] T ,
G m = [ g n − 1 g n − 2 g n ] T , G m + 1 = [ g n + u g n + v g n + w ] T
θ 11 = u [ 5 u 5 v + 5 u 5 w + 14 u 3 v 3 − 16 u 4 v 2 + 14 u 3 w 3 − 16 u 4 w 2 + 210 v 3 w 3 − 23 u 4 v w − 56 u v 2 w 3 − 56 u v 3 w 2 − 28 u 2 v w 3 − 28 u 2 v 3 w + 40 u 3 v w 2 + 40 u 3 v 2 w ] 420 v 3 w 3
θ 12 = u [ − 350 u 5 v − 350 u 5 w − 140 u 3 v 3 + 388 u 4 v 2 − 140 u 3 w 3 + 388 u 4 w 2 + 210 v 3 w 3 + 105 u 6 + 1554 u 2 v 2 w 2 + 1187 u 4 v w − 574 u v 2 w 3 − 574 u v 3 w 2 + 490 u 2 v w 3 + 490 u 2 v 3 w − 1342 u 3 v w 2 − 1342 u 3 v 2 w ] 420 ( u − w ) 3 ( u − v ) 3
θ 13 = − u 5 [ 10 u 4 v − 5 u 4 w + 28 u 2 v 3 − 35 u 3 v 2 − 14 u 2 w 3 + 16 u 3 w 2 − 70 v 2 w 3 + 98 v 3 w 2 + 70 u v w 3 − 98 u v 3 w − 10 u 3 v w − 42 u v 2 w 2 − 46 u 2 v w 2 + 98 u 2 v 2 w ] 420 v 3 ( v − w ) 3 ( u − v ) 3
θ 14 = − u 5 [ 5 u 4 v − 10 u 4 w + 14 u 2 v 3 − 16 u 3 v 2 − 28 u 2 w 3 + 35 u 3 w 2 3 − 98 v 2 w + 70 v 3 w 2 + 98 u v w 3 − 70 u v 3 w + 10 u 3 v w + 42 u v 2 w 2 − 98 u 2 v w 2 + 46 u 2 v 2 w ] 420 w 3 ( v − w ) 3 ( u − w ) 3
θ 15 = u 2 [ − 16 u 3 v − 16 u 3 w + 14 u 2 v 2 + 14 u 2 w 2 + 70 v 2 w 2 + 5 u 4 − 56 u v w 2 − 56 u v 2 w + 56 u 2 v w ] 840 v 2 w 2
θ 16 = − u 2 [ − 40 u 3 v − 40 u 3 w + 28 u 2 v 2 + 28 u 2 w 2 + 70 v 2 w 2 + 15 u 4 − 84 u v w 2 − 84 u v 2 w + 112 u 2 v w ] 840 ( u − w ) 2 ( u − v ) 2
θ 17 = u 5 − 8 u 2 v + 14 u w 2 − 16 u 2 w − 28 v w 2 + 5 u 3 + 28 u v w 840 v 2 ( v − w ) 2 ( u − v ) 2
θ 18 = u 5 14 u v 2 − 16 u 2 v − 8 u 2 w − 28 v 2 w + 5 u 3 + 28 u v w 840 w 2 ( v − w ) 2 ( u − w ) 2
θ 21 = v [ 5 u v 5 + 5 v 5 w − 16 u 2 v 4 + 14 u 3 v 3 + 210 u 3 w 3 + 14 v 3 w 3 − 16 v 4 w 2 − 23 u v 4 w − 28 u v 2 w 3 + 40 u v 3 w 2 − 56 u 2 v w 3 + 40 u 2 v 3 w − 56 u 3 v w 2 − 28 u 3 v 2 w ] 420 u 3 w 3
θ 22 = v 5 [ 10 u v 4 − 5 v 4 w − 35 u 2 v 3 + 28 u 3 v 2 − 70 u 2 w 3 + 98 u 3 w 2 − 14 v 2 w 3 + 16 v 3 w 2 + 70 u v w 3 − 10 u v 3 w − 98 u 3 v w − 46 u v 2 w 2 − 42 u 2 v w 2 + 98 u 2 v 2 w ] 420 u 3 ( u − w ) 3 ( u − v ) 3
θ 23 = v [ 350 u v 5 + 350 v 5 w − 388 u 2 v 4 + 140 u 3 v 3 − 210 u 3 w 3 + 140 v 3 w 3 − 388 v 4 w 2 − 105 v 6 − 1554 u 2 v 2 w 2 − 1187 u v 4 w − 490 u v 2 w 3 + 1342 u v 3 w 2 + 574 u 2 v w 3 + 1342 u 2 v 3 w + 574 u 3 v w 2 − 490 u 3 v 2 w ] 420 ( v − w ) 3 ( u − v ) 3
θ 24 = − v 5 [ 5 u v 4 − 10 v 4 w − 16 u 2 v 3 + 14 u 3 v 2 − 98 u 2 w 3 + 70 u 3 w 2 − 28 v 2 w 3 + 35 v 3 w 2 + 98 u v w 3 + 10 u v 3 w − 70 u 3 v w − 98 u v 2 w 2 + 42 u 2 v w 2 + 46 u 2 v 2 w ] 420 w 3 ( v − w ) 3 ( u − w ) 3
θ 25 = v 2 [ − 16 u v 3 − 16 v 3 w + 14 u 2 v 2 + 70 u 2 w 2 + 14 v 2 w 2 + 5 v 4 − 56 u v w 2 + 56 u v 2 w − 56 u 2 v w ] 840 u 2 w 2
θ 26 = − v 5 8 u v 2 + 28 u w 2 − 14 v w 2 + 16 v 2 w − 5 v 3 − 28 u v w 840 u 2 ( u − w ) 2 ( u − v ) 2
θ 27 = − v 2 [ − 40 u v 3 − 40 v 3 w + 28 u 2 v 2 + 70 u 2 w 2 + 28 v 2 w 2 + 15 v 4 − 84 u v w 2 + 112 u v 2 w − 84 u 2 v w ] 840 ( v − w ) 2 ( u − v ) 2
θ 28 = v 5 − 16 u v 2 + 14 u 2 v − 28 u 2 w − 8 v 2 w + 5 v 3 + 28 u v w 840 w 2 ( v − w ) 2 ( u − w ) 2
θ 31 = w [ 5 u w 5 + 5 v w 5 + 210 u 3 v 3 − 16 u 2 w 4 + 14 u 3 w 3 − 16 v 2 w 4 + 14 v 3 w 3 − 23 u v w 4 + 40 u v 2 w 3 − 28 u v 3 w 2 + 40 u 2 v w 3 − 56 u 2 v 3 w − 28 u 3 v w 2 − 56 u 3 v 2 w ] 420 u 3 v 3
θ 32 = w 5 [ 10 u w 4 − 5 v w 4 − 70 u 2 v 3 + 98 u 3 v 2 − 35 u 2 w 3 + 28 u 3 w 2 + 16 v 2 w 3 − 14 v 3 w 2 − 10 u v w 3 + 70 u v 3 w − 98 u 3 v w − 46 u v 2 w 2 + 98 u 2 v w 2 − 42 u 2 v 2 w ] 420 u 3 ( u − w ) 3 ( u − v ) 3
θ 33 = w 5 [ 5 u w 4 − 10 v w 4 − 98 u 2 v 3 + 70 u 3 v 2 − 16 u 2 w 3 + 14 u 3 w 2 + 35 v 2 w 3 − 28 v 3 w 2 + 10 u v w 3 + 98 u v 3 w − 70 u 3 v w − 98 u v 2 w 2 + 46 u 2 v w 2 + 42 u 2 v 2 w ] 420 v 3 ( v − w ) 3 ( u − v ) 3
θ 34 = w [ − 350 u w 5 − 350 v w 5 + 210 u 3 v 3 + 388 u 2 w 4 − 140 u 3 w 3 + 388 v 2 w 4 − 140 v 3 w 3 + 105 w 6 + 1554 u 2 v 2 w 2 + 1187 u v w 4 − 1342 u v 2 w 3 + 490 u v 3 w 2 − 1342 u 2 v w 3 − 574 u 2 v 3 w + 490 u 3 v w 2 − 574 u 3 v 2 w ] 420 ( v − w ) 3 ( u − w ) 3
θ 35 = w 2 [ − 16 u w 3 − 16 v w 3 + 70 u 2 v 2 + 14 u 2 w 2 + 14 v 2 w 2 + 5 w 4 + 56 u v w 2 − 56 u v 2 w − 56 u 2 v w ] 840 u 2 v 2
θ 36 = − w 5 28 u v 2 + 8 u w 2 + 16 v w 2 − 14 v 2 w − 5 w 3 − 28 u v w 840 u 2 ( u − w ) 2 ( u − v ) 2
θ 37 = − w 5 28 u 2 v + 16 u w 2 − 14 u 2 w + 8 v w 2 − 5 w 3 − 28 u v w 840 v 2 ( v − w ) 2 ( u − v ) 2
θ 38 = − w 2 [ − 40 u w 3 − 40 v w 3 + 70 u 2 v 2 + 28 u 2 w 2 + 28 v 2 w 2 + 15 w 4 + 112 u v w 2 − 84 u v 2 w − 84 u 2 v w ] 840 ( v − w ) 2 ( u − w ) 2
For our methods, in case 1, we considered one step method with two hybrid
points with equal interval where u = 1 3 , v = 2 3 , w = 1 . For case II, we considered
two step method with one hybrid points with equal interval, where u = 1 , v = 3 / 2 , w = 2 and for the lase case III, we considered three step method with equal interval where u = 1 , v = 2 , w = 3 .
In this section, we investigate the basic properties of the developed method vis-a-vis order, local truncation error, consistency, zero-stability, convergence, and region of absolute stability of the methods.
The operation l is associated with the linear method defined by
l [ y ( x ) : h ] = ζ ( 1 ) Y m + 1 − ζ ( 0 ) Y m − h ( η ( 0 ) F m + η ( 1 ) F m + 1 ) − h 2 ( γ ( 0 ) G m + γ ( 1 ) G m + 1 ) (9)
where y ( x ) is an arbitrary function, continuously differentiable on an interval [ x n , x N ] . Ehigie, J. O. and Okunuga [
l [ y ( x ) : h ] = c 0 y ( x n ) + c 1 h y ' ( x n ) + c 2 h 2 y ″ ( x n ) + ⋯ + c q h q y ( q ) ( x n ) + ⋯
where
c p = 1 p [ ∑ j = 1 r j p θ j − 1 ( p − 1 ) ! ∑ j = 1 r j p − 1 γ j ]
(9) is of order p if
l [ y ( x ) : h ] = o ( h p + 1 ) , c 0 = c 1 = ⋯ = c p ≠ c p + 1 = 0
c p + 1 is called the error constant and c p + 1 h p + 1 y ( p + 1 ) ( x ) is called the local truncation error (LTE).
For our scheme, the order is 8 with error constant given by
Definition 1 A block method is consistent if it has order p ≥ 1.
Definition 2 A method is said to be Zero-stable if no root of the first characteristics polynomial has modulus greater than one, and if every root of modulus one has multiplicity not greater than one or is simple.
ρ ( λ ) = | λ [ 1 0 0 0 1 0 0 0 1 ] − [ 0 0 1 0 0 1 0 0 1 ] |
The roots of the determinant gives λ = 0 , 0 , 1 , hence the method is consistent.
Definition 3 A method is said to be convergent if It is consistent and zero stable.
Definition 4 The Region of absolute stability (RAS) of a L M M is the set R = { z = λ h : for z where the root of the stability polynomial are absolute less than one}.
Substituting the test equation y ′ = λ y , y ″ = λ 2 y and writing r = e i θ , the region of absolute stability for our new methods are shown below:
In determining the region of absolute stability, we consider three cases in this paper as described below:
Definition 5 A-Stability
A method is said to be A-stable if its region of absolute stability contains the whole of the complex left hand-half plane Re ( h λ ) < 0 . Alternatively, a numerical method is called A-Stable if the solution tend to zero as n → ∞ when the method is applied with fixed h to any differential equation of the form
∂ y ∂ x = λ y , where l is a complex constant with negative real part. Hence, for our cases I, II, & III; The regions of absolute stabilities are given in Figures 1-3.
We conclude that the three cases presented in this paper are A-stable.
In this section we considered four examples to test the efficiency of the method. We compared the results of the cases in order to conclude on the best way to fix u,v and w. The following notations are used to show the results ψ ψ e − υ υ = ψ ψ ∗ 10 − υ υ .
Example 1 We consider a system in the range 0 ≤ x ≤ 20
y ′ = ( − 1 95 − 1 − 97 ) y , y ( x ) = (11)
with the exact solution
y ( x ) = 1 47 ( 95 e − 2 x − 48 e − 95 x 48 e − 96 x − e − 2 x )
The eigenvalues of the Jacobian matrix are λ 1 = − 2 , λ 2 = − 96 with the stiffness ratio 1:48. Source :Abdulimen [
Example 2 We consider a simple nonlinear stiff system
y ′ 1 = − y 1 , y 1 ( 0 ) = 5 y ′ 2 = y 1 2 − 2 y 2 , y 2 ( 0 ) = 5
x ∈ [ 0,20 ] , with the exact solution
( y 1 y 2 ) = ( 5 e x p ( − x ) 5 e x p ( − 2 x ) ( 1 _ 5 x ) )
The eigenvalues of the Jacobian matrix are λ 1 = − 1 , λ 2 = − 2 . Source: Yakubu and Markus [
Example 3 In the this example we consider stiff nonlinear system of two dimensional Kaps problem with corresponding initial conditions
[ y ′ 1 ( x ) y ′ 2 ( x ) ] = [ − 1002 y 1 ( x ) + 1000 y 2 ( x ) 2 y 1 ( x ) − y ( x ) ( 1 + y 2 ( x ) ) ] , [ y 1 ( 0 ) y 2 ( 0 ) ] = [ 1 1 ]
The exact solution is
[ y 1 ( x ) y 2 ( x ) ] = [ e x p ( − 2 x ) e x p ( − x ) ]
Source: Yakubu and Markus [
Example 4 We consider a four dimensional problems
( y ′ 1 ( x ) y ′ 2 ( x ) y ′ 3 ( x ) y ′ 4 ( x ) ) = ( − 10 4 y 1 ( x ) + 100 y 2 ( x ) − 10 y 3 ( x ) + y 4 ( x ) − 1000 y 2 ( x ) + 10 y 3 ( x ) − 10 y 4 ( x ) − y 3 ( x ) + 10 y 4 ( x ) − 0.1 y 4 ( x ) ) = ( y 1 ( 0 ) y 2 ( 0 ) y 3 ( 0 ) y 4 ( 0 ) ) = ( 1 1 1 1 )
Within the range 0 ≤ x ≤ 20 . The eigenvalues of the Jacobian matrix λ 1 = − 0.1 , λ 2 = − 10 , λ 3 = − 1000 and λ 4 = − 10000 . The exact solution is given as
y 1 ( x ) = − 89990090 8999010009 e − 0.1 x + 818090 89901009 e − x + 998911 899010090 e − 1000 x + 89071119179 89990100090 e − 10000 x
y 2 ( x ) = 9100 8991 e − 0.1 x − 910 8991 e − x + 9989911 9989001 e − 1000 x
y 3 ( x ) = 100 9 e − 0.1 x − 91 9 e − x
y 4 ( x ) = e − 0.1 x
Source: Ehigie, Okunuga, and Sofoluwe [
In this paper, we introduced three new second derivative linear multistep methods for the numerical solution of stiff initial value problems. Four numerical examples were considered, the results justified the proficiency of the second derivative method which is cheaper to implement since it does not require starting values and particularly the new methods show that lower step method gives better accuracy than higher step methods of the same order. We are able to achieve the aim of this paper which is to develop a class of second derivative linear multistep methods that are A-stable with large region of absolute stabilityas shown in Figures 1-3. The results from the high-order methods are very encourging (see Tables 1-4), therefore, we recommend further investigation of the second-derivative.
Adewale, J., Olaide, A., Oziohu, O., Joshua, S. and Omuya, M. (2018) Proficiency of Second Derivative Schemes for the Numerical Solution of Stiff Systems. American Journal of Computational Mathematics, 8, 96-107. https://doi.org/10.4236/ajcm.2018.81008