Mathematical
and Computational
Applications
Article
Direct Solution of Second-Order Ordinary
Differential Equation Using a Single-Step Hybrid
Block Method of Order Five
Ra’ft Abdelrahim *,† and Zurni Omar †
Department of mathematics, College of Art and Sciences, Universiti Utara Sintok, Kedah 0060, Malaysia;
[email protected]
* Correspondence: [email protected]; Tel.: +60-11-1150-3218
† These authors contributed equally to this work.
Academic Editor: Mehmet Pakdemirli
Received: 1 February 2016; Accepted: 18 March 2016; Published: 12 April 2016
Abstract: This paper proposes a new hybrid block method of order five for solving second-order
ordinary differential equations directly. The method is developed using interpolation and collocation
techniques. The use of the power series approximate solution as an interpolation polynomial and
its second derivative as a collocation equation is considered in deriving the method. Properties of
the method such as zero stability, order, consistency, convergence and region of absolute stability
are investigated. The new method is then applied to solve the system of second-order ordinary
differential equations and the accuracy is better when compared with the existing methods in terms
of error.
Keywords: single-step; hybrid block method; system of second order ordinary differential equations;
collocation and Interpolation method; direct solution
Subject Classificatio: 65L05; 65L06; 65L20
1. Introduction
Ordinary differential equations (ODEs) are commonly used for mathematical modeling in many
diverse fields such as engineering, operation research, industrial mathematics, behavioral sciences,
artificial intelligence, management and sociology. This mathematical modeling is the art of translating
problems from an application area into tractable mathematical formulations whose theoretical and
numerical analysis provide insight, answers and guidance useful for the originating application [1].
This type of problem can be formulated either in terms of first-order or higher-order ODEs.
In this article, the system of second-order ODEs of the following form is considered.
1 y2
“ 1 f px, 1 y, 1 y1 q,
2 y2 “ 2 f px, 2 y, 2 y1 q,
1 ypx
“ a0 , 1 y1 px0 q “ b0
2 ypx q “ a , 2 y1 px q “ b
0
0
1
1
..
.
m y2 “ m f px, m y, m y1 q, m ypx q “ a , m y1 px q “ b
n
n
0
0
0q
(1)
The method of solving higher-order ODEs by reducing them to a system of first-order approach
involves more functions to evaluate them and then leads to a computational burden as mentioned
in [2–4]. The multistep methods for solving higher-order ODEs directly have been developed by many
scholars such as [5–9]. However, these researchers only applied their methods to solve single initial
value problems of ODEs.
Math. Comput. Appl. 2016, 21, 12; doi:10.3390/mca21020012
www.mdpi.com/journal/mca
Math. Comput. Appl. 2016, 21, 12
2 of 7
The aim of this paper is to develop a new numerical method for solving single second-order ODEs
and systems of second-order ODEs directly.
2. Derivation of the Method
In this section, a one-step hybrid block method with three off-step points, xn` 1 , xn` 2 and xn` 3 ,
5
5
5
for solving Equation (1) is derived. Let the power series of the form
j
ypxq “
v`ÿ
m ´1
ˆ
ai
i “0
x ´ xn
h
˙i
j “ 1, . . . , m.
,
(2)
be the approximate solution to Equation (1) for x P rxn , xn`1 s where n “ 0, 1, 2, ¨ ¨ ¨ , N ´ 1, a 1s
are the real coefficients to be determined, v is the number of collocation points, m is the number of
interpolation points and h “ xn ´ xn´1 is a constant step size of the partition of interval ra, bs which is
given by a “ x0 ă x1 ă ¨ ¨ ¨ ă x N ´1 ă x N “ b.
Differentiating Equation (2) twice gives:
j
j
2
j
j 1
y pxq “ f px, y, y q “
v`ÿ
m ´1
i “2
ipi ´ 1qai
h2
ˆ
x ´ xn
h
˙ i ´2
,
j “ 1, . . . , m.
(3)
Interpolating Equation (2) at xn` 2 , xn` 3 and collocating Equation (3) at all points in the selected
5
5
interval, i.e., xn , xn` 1 , xn` 2 , xn` 3 and xn` 1 , gives the following equations which can be written in
5
5
5
matrix form:
¨
˛
˛¨
˛ ¨ j
4
8
16
32
64
y n` 2
1 25 25
a0
125
625
3125
15625
5
˚
‹
‹ ˚
9
27
81
243
729 ‹ ˚
˚ 1 53 25
‹ ˚ a 1 ‹ ˚ j y n` 3 ‹
125
625
3125
15625
5 ‹
˚
‹˚
‹ ˚
˚ 0 0 2
‹
˚
‹ ˚ jf
0
0
0
0 ‹
n
˚
‹
‹ ˚ a2 ‹ ˚
h2
˚
‹
‹
˚
‹
˚
6
12
4
6
jf
‹
˚ 0 0 22
‹
˚
‹
˚
a
1
(4)
“
h
5h2
25h2
25h2
125h2 ‹ ˚ 3 ‹
˚
˚ n` 5 ‹ , j “ 1, . . . , m.
˚
‹
‹˚
‹ ˚ j
12
48
32
96
2
˚ 0 0 h2 5h2 25h2 25h2 125h2 ‹ ˚ a4 ‹ ˚ f n` 2 ‹
5 ‹
˚
‹˚
‹ ˚
18
108
108
486 ‹ ˚
˚ 0 0 2
a ‹ ˚ j f n` 3 ‹
‚
˝
h2
5h2
25h2
25h2
125h2 ‚˝ 5 ‚ ˝
5
6
12
20
30
jf
0 0 h22
a
6
n
`
1
h2
h2
h2
h2
Employing the Gaussian elimination method on Equation (4) gives the coefficient ai 's, for i “ 0p1q6.
These values are then substituted into Equation (2) to give the implicit continuous hybrid method of
the form:
j
ÿ
y pxq “
j
α i pxq j y n`i `
i“ 52 , 35
ÿ
j
β i pxq j f n`i `
i“ 51 , 25 , 53
1
ÿ
j
β i pxq j f n`i ,
j “ 1, . . . , m..
(5)
i “0
Differentiating Equation (5) once yields:
j 1
y pxq “
ÿ
i“ 25 , 35
d j
α pxq j y n`i `
dx i
ÿ
i“ 51 , 25 , 53
where
j
1
ÿ d
d j
j
β i pxq j f n`i `
β pxq j f n`i , j “ 1, . . . , m.
dx
dx i
i “0
ˆ
5px ´ xn q
´2
h
˙
ˆ
5px ´ xn q
3´
h
˙
α n` 3 “
5
j
α n` 2 “
5
(6)
Math. Comput. Appl. 2016, 21, 12
j
px ´ xn q2 61px ´ xn q3 205px ´ xn q4 55px ´ xn q5 25px ´ xn q6 301hpx ´ xn q 11h2
´
`
´
`
`
´
2
36h
4500
3750
72h2
24h3
36h4
β0 “
j
25px ´ xn q3 775px ´ xn q4 125px ´ xn q5 125px ´ xn q6 25hpx ´ xn q 83h2
`
´
´
´
`
8h
96
2000
96h2
16h3
48h4
β1 “
5
β2 “
575px ´ xn q4 25px ´ xn q3 75px ´ xn q5 125px ´ xn q6 43hpx ´ xn q 17h2
´
´
`
´
`
12h
300
250
72h2
8h3
36h4
β3 “
25px ´ xn q3 425px ´ xn q4 25px ´ xn q5 125px ´ xn q6 109hpx ´ xn q 23h2
`
´
´
´
`
36h
3600
3000
144h2
6h3
72h4
j
55px ´ xn q4 px ´ xn q3 5px ´ xn q5 25px ´ xn q6 11hpx ´ xn q
h2
´
`
`
´
´
2
3
4
24h
12000
10000
288h
16h
144h
j
5
j
3 of 7
5
β1 “
Equation (5) is evaluated at non-interpolating points, i.e., xn , xn` 1 and xn+1 , while its first
5
derivative is evaluated at all points, and this yields the following equation in matrix form:
j
A j Y L “ j B j R 1 ` j C j R 2 ` j D j R 3 j “ 1, . . . , m
(7)
where
¨
¨
˚
˚
˚
˚
˚
˚
j
A“˚
˚
˚
˚
˚
˚
˝
0
1
0
0
0
0
0
0
¨
˚
˚
˚
˚
˚
˚
˚
˚
C“˚
˚
˚
˚
˚
˚
˚
˚
˝
´3
´2
2
2
1
´3
5
h
5
h
5
h
5
h
5
h
´5
h
´5
h
´5
h
´5
h
´5
h
11h2
3750
´ h2
7500
´21h2
2500
´301h
4500
7h
1800
´h
500
19h
9000
´53h
1000
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
˛
5
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹ j
˚
‹ , YL “ ˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
˚
‚
˝
y n` 2
5
y n` 3
5
y n `1
y 1 n` 1
5
y 1 n` 2
5
y 1 n` 3
5
y 1 n `1
¨
˛
y n` 1
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
´ ¯
˚
‹ j
‹ , R2 “ j f n j D “ ˚
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˝
‚
83h2
2000
23h2
6000
83h2
2000
´25h
96
´69h
800
31h
2400
´31h
2400
25h
96
17h2
250
49h2
1500
´113h2
1500
´43h
300
´371h
1800
´77h
900
31h
600
´883h
1800
˛
˛
¨
´1 0
‹
‹
˚ 0
0 ‹
‹
‹
˚
‹
‹
˚
‹
˚ 0
0 ‹
˜
¸
‹
‹
˚
‹ j
˚ 0 ´1 ‹
y
n
‹ , R1 “
‹, B “ ˚
‹
˚ 0
y1n
0 ‹
‹
‹
˚
‹
‹
˚
0 ‹
‹
˚ 0
‹
‹
˚
‹
˝ 0
0 ‚
‚
0
0
23h2
3000
11h2
3000
151h2
1000
´109h
3600
´41h
3600
´31
1200
43h
720
797h
1200
´ h2
10000
´ h2
30000
337h2
30000
11h
12000
´h
7200
17h
36000
´7h
12000
4283h
36000
˛
‹
‹
‹
‹
¨ j
‹
f n` 1
‹
5
‹
˚ j
‹ j
˚ f n` 2
5
‹ , R3 “ ˚
‹
˚ jf 3
‹
˝ n` 5
‹
jf
‹
n `1
‹
‹
‹
‚
˛
‹
‹
‹
‹
‚
Multiplying Equation (7) by j A ´ 1 gives the hybrid block method as shown below.
I jY L “ j B j R1 ` j C j R2 ` j D j R3
¨
¨
˚
˚
˚
˚
˚
˚
I“˚
˚
˚
˚
˚
˚
˝
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
¨
1
˚
‹
˚
1
‹
˚
‹
˚ 1
‹
˚
‹
˚
‹ j
‹, B “ ˚
˚ 1
‹
˚ 0
‹
˚
‹
˚
‹
˚ 0
‹
˚
‚
˝ 0
0
˛
h
5
2h
5
3h
5
h
1
1
1
1
˛
˚
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹ j
‹, C “ ˚
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‚
˚
˝
58h2
5625
134h2
2500
93h2
2500
h2
18
637h
9000
73h
1125
69h
1000
h
72
˛
(8)
¨
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹ j
˚
‹, D “ ˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‹
˚
‚
˝
173h2
12000
47h2
750
459h2
4000
25h 2
96
209h
1200
41h
150
99h
400
25h
48
´ h2
150
´4h2
375
9h2
500
0
´113h
1800
13h
225
39h
200
´25h
72
37h2
18000
h2
225
21h2
2000
25h2
144
17h
900
h
225
9h
100
25h
36
´7h2
60000
´ h2
3750
´9h2
20000
h2
96
´19h
18000
´h
2250
´3h
2000
17h
144
˛
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‚
Math. Comput. Appl. 2016, 21, 12
4 of 7
3. Properties of the Method
3.1. Zero Stability
The one-step hybrid block method (8) is said to be zero-stable if and only if the first characteristic
function Π(x) has roots such that |xt | ď 1 and if |xt | “ 1; then the multiplicity of xt does not exceed
two. The characteristic function of the new derived method is given as below:
¨
˚
˚
˚
˚
˚
˚
Πpxq “ ˚
˚
˚
˚
˚
˚
˝
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
¨
0
0
0
1
0
0
0
‹ ˚
‹ ˚
‹ ˚
‹ ˚
‹ ˚
‹ ˚
‹´˚
‹ ˚
‹ ˚
‹ ˚
‹ ˚
‹ ˚
‚ ˚
˝
0
0
0
1
0
0
0
0
0
0
1
0
0
0
h
5
2h
5
3h
5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
˛
0
0
0
0
0
0
0
1
˛
‹
‹
‹
‹
‹
‹
‹
2
‹ “ x6 px ´ 1q
‹
‹
‹
‹
‹
‚
the solution of which is x “ 0, 0, 0, 0, 0, 0, 1, 1. Hence, our method is zero-stable.
3.2. Order of the Method
According to [9] the order of the new method in Equation (8) is obtained by using the Taylor series
and it is found that the developed method has an order of r5, 5, 5, 5, 5, 5, 5, 5sT with an error constant
vector of:
[1.732063 ˆ 10´7 , 4.104127 ˆ 10´7 , 6.582857 ˆ 10´7 , ´1.587302 ˆ 10´6 , 1.537778 ˆ 10´6 ,
8.533333 ˆ 10´7 , 1.680000 ˆ 10´6 , ´1.666667 ˆ 10´5 ]T
3.3. Consistency
The hybrid block method is said to be consistent if it has an order more than or equal to one.
Therefore, our method is consistent.
3.4. Convergence
Zero stability and consistency are sufficient conditions for a linear multistep method to be
convergent [10]. Since the new hybrid block method is zero-stable and consistent, it can be concluded
that the method is convergent.
3.5. Region of Absolute Stability
In this section, the locus boundary method is adopted to determine the region of absolute stability.
The linear multistep numerical method is said to be absolutely stable if for all given h, the roots of the
characteristic function Πpxq “ ρpxq ´ hσpxq satisfies |x| ă 1. The test equation y “ ´λ2 y is substituted
df
in Equation (8) where h “ λ2 h2 and λ “ dy . Substituting x “ cos θ ´ isinθ and equating the imaginary
part yields:
p56250000 pcos pθq ´ 1qq
h“
p3cos pθq ´ 89q
This gives the stability interval of (0, 1222826).
4. Implementation of the Method
The initial starting value at each block is obtained by using the Taylor method. Then, the
calculations are corrected using Equation (8). For the next block, the same techniques are repeated to
compute the approximation values of j y n` 1 , j y n` 2 , j y n` 3 , j y n` 1 , j = 1, . . . , m simultaneously
5
5
5
Math. Comput. Appl. 2016, 21, 12
5 of 7
until the end of the integrated interval. During the calculations of the iteration, the final values of
jy
n`1 are taken as the initial values for the next iteration.
5. Numerical Experiments
In this section, the performance of the developed one-step hybrid block method is examined using
the following three systems of second-order initial value problems. Tables 1 and 2 Tables 3 and 4 and
Tables 5 and 6 show the comparison of the numerical results of the new method with exact solution for
solving problems 1–3 respectively. While, in Table 7, the results of the developed method are more
accurate than that of [11] which was executed by six-step block method for solving Problem 4.
2
Problem 1: y1 “ 1 ´ cosx ` sinpy12 q ` cospy12 q y1 p0q “ 1, y11 p0q “ 0,
2
1
y2 “ p4`1y2 q ´
y2 p0q “ 0, y12 p0q “ π,
p5´sinpxq2 q
1
Exact solution: y1 = cos x, y2 = πx
Table 1. Exact solution and computed solution of the new method for solving y1 in Problem 1.
x
Exact Solution of y1
Computed Solution of y1
Error in y1
0.2
0.4
0.6
0.8
1.0
0.98006657784124163
0.92106099400288510
0.82533561490967833
0.69670670934716550
0.54030230586813977
0.98006661117494787
0.92106172165508671
0.82533871008804471
0.69671472046421035
0.54031839116566260
3.333371 ˆ 10´8
7.276522 ˆ 10´7
3.095178 ˆ 10´6
8.011117 ˆ 10´6
1.608530 ˆ 10´5
Table 2. Exact solution and computed solution of the new method for solving y2 in Problem 1.
x
Exact solution of y2
Computed solution of y2
Error in y2
0.2
0.4
0.6
0.8
1.0
0.62831853071795862
1.25663706143591720
1.88495559215387590
2.51327412287183450
3.14159265358979270
0.62831853071222155
1.25663706109624340
1.88495558867424440
2.51327410626325070
3.14159260122723750
5.737077 ˆ 10´12
3.396738 ˆ 10´10
3.479631 ˆ 10´9
1.660858 ˆ 10´8
5.236256 ˆ 10´8
2
Problem 2: y1 “ ´e´x y2 y1 p0q “ 1, y11 p0q “ 0, h “ 0.01
2
y2 “ 2e x y11 y2 p0q “ 1, y12 p0q “ 1,
Exact solution: y1 = cos x, y2 = ex cos x
Table 3. Exact solution and computed solution of the new method for solving y1 in Problem 2.
x
Exact Solution of y1
Computed Solution of y1
Error in y1
0.2
0.4
0.6
0.8
1.0
0.980066577841241630
0.921060994002884990
0.825335614909678110
0.69670670934716505
0.540302305868139210
0.980066574492776010
0.921060961237438300
0.825335481688297850
0.696706354719187630
0.540301570350463110
3.348466 ˆ 10´9
3.276545 ˆ 10´8
1.332214 ˆ 10´7
3.546280 ˆ 10´7
7.355177 ˆ 10´7
Math. Comput. Appl. 2016, 21, 12
6 of 7
Table 4. Exact solution and computed solution of the new method for solving y2 in Problem 2.
x
Exact Solution of y2
Computed Solution of y2
Error in y2
0.2
0.4
0.6
0.8
1.0
1.197056021355891400
1.374061538887522100
1.503859540558786200
1.550549296807422400
1.468693939915884900
1.197056651769760100
1.374064060556476500
1.503864970151431300
1.550558149588110000
1.468705886868917300
6.304139 ˆ 10´7
2.521669 ˆ 10´6
5.429593 ˆ 10´6
8.852781 ˆ 10´8
1.194695 ˆ 10´8
´y
2
Problem 3: y1 “ b 1
y21 `y22
´y
2
y2 “ b 2
y21 `y22
y1 p0q “ 1, y11 p0q “ 0, h “ 0.01
y2 p0q “ 0, y12 p0q “ 1,
Exact solution: y1 = cos x, y2 = sin x
Table 5. Exact solution and computed solution of the new method for solving y1 in Problem 3.
x
Exact Solution of y1
Computed Solution of y1
Error in y1
0.2
0.4
0.6
0.8
1.0
0.980066577841241630
0.921060994002884990
0.825335614909678110
0.696706709347165050
0.540302305868139210
0.980066577799155510
0.921060993708316070
0.825335614150025320
0.696706708168561060
0.540302304687844350
4.208611e´11
2.945689e´10
7.596528e´10
1.178604e´9
1.180295e´9
Table 6. Exact solution and computed solution of the new method for solving y2 in Problem 3.
x
Exact Solution of y2
Computed Solution of y2
Error in y2
0.2
0.4
0.6
0.8
1.0
0.198669330795061240
0.389418342308650690
0.564642473395035590
0.717356090899523120
0.841470984807896840
0.198669331113754400
0.389418343391428780
0.564642475168185330
0.717356092762203360
0.841470985889528840
3.186932 ˆ 10´10
1.082778 ˆ 10´9
1.773150 ˆ 10´9
1.862680 ˆ 10´9
1.081632 ˆ 10´9
` ˘2
Problem 4: y2 “ x y1 , y p0q “ 1, y1 p0q “ 12 , h “
´
¯
Exact solution y “ 1 ` 21 ln 22`_xx
1
30 .
Table 7. Comparison of the new method with [11] for solving Problem 4.
x
Exact Solution
Computed Solution
Error in New
Method P = 5
Error in [11] P = 7
0.1
0.2
0.3
0.4
1.0
1.0500417292784914
1.1003353477310756
1.1511404359364668
1.2027325540540821
1.2554128118829952
1.0500417292785045
1.1003353477311153
1.1511404359364565
1.2027325540537517
1.2554128118817025
1.310063 ˆ 10´16
3.974598 ˆ 10´14
1.021405 ˆ 10´14
3.304024 ˆ 10´13
1.292744 ˆ 10´12
1.445510 ˆ 10´14
3.779332 ˆ 10´13
3.428134 ˆ 10´11
6.987109 ˆ 10´8
2.017066 ˆ 10´7
The numerical results confirm that the proposed method produces better accuracy if compared
with the existing methods. This is also clear in the graph below (Figure 1).
0.4
1.0
1.2027325540540821
1.2554128118829952
1.2027325540537517
1.2554128118817025
3.304024 × 10−13
1.292744 × 10−12
6.987109 × 10−8
2.017066 × 10−7
The numerical results confirm that the proposed method produces better accuracy if compared
with
existing
methods.
Math. the
Comput.
Appl. 2016,
21, 12 This is also clear in the graph below (Figure 1).
7 of 7
Figure 1. Comparison between errors in the new method with error in [11] for solving Problem 4.
Figure 1. Comparison between errors in the new method with error in [11] for solving Problem 4.
6. Conclusions
Conclusions
6.
In this
this article,
article, aa one-step
one-step block
block method
method with
with three
three off-step
off-step points
points is
is derived
derived via
via the
the interpolation
interpolation
In
collocation approach.
approach. The
The developed
developed method
method is
is consistent,
consistent, zero-stable,
zero-stable, convergent,
convergent, with
with aa region
region of
of
collocation
absolute
stability
and
order
five.
The
numerical
results
generated
when
the
new
developed
method
absolute stability and order five. The numerical results generated when the new developed method
was applied
applied to
to three
three systems
systems of
of second-order
second-order initial
initial value
value problems
problems above
above have
have shown
shown the
the high
high
was
accuracy of
of the
the new
new method.
method.
accuracy
Author Contributions: Both authors equally contributed to this work.
Author Contributions: Both authors equally contributed to this work.
Conflicts of Interest: The authors declare no conflict of interest.
Conflicts of Interest: The authors declare no conflict of interest.
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