1
Numerical Integration Formulas for Solving the
Initial Value Problem of Ordinary Differential Equations
Maitree Podisuk and Wannaporn Sanprasert
Department of Mathematics and Computer Science
King Mongkut’s Institute of Technology Chaokhuntaharn Ladkrabang
Ladkrabang Bangkok 10520
THAILAND
Abstract: In this paper, we will use four numerical integration formulas to solve the
initial value problem of the ordinary differential equations by solving the integral
equation instead of the ordinary differential equation. We will use these four formulas
to find the numerical solutions of some examples and compare these results with the
known Runge-Kutta formulas, Euler’s formula, Goeken-Johnson formula and Wu’s
formula.
Key-Words: Runge-Kutta Midpoint Trapezoidal Simpson Goeken Johnson Wu
1
Introduction
The initial value problem of the
ordinary differential equation is of the
form
y ( x )  f ( x, y), x  [a , b]
(1)
with the initial condition
y (a )  c .
(2)
If we divide the close interval [a , b]
into n subintervals at the points
x 0  a  x1  x 2  ...  x n  b , where
h  x11  x i for i  0,1,2,..., n then
some known numerical formulas for
finding the numerical solution of the
above equations at these points are
(3)
y m1  y m  hf (x m , y m )
which is Euler formula which we shall
denote by RK1,
h
y m1  y m  K1  K 2 
(4)
2
where K1  f (x m , y m )
K 2  f (x m  h, y m  hK 1 )
which is two points Runge-Kutta
formula which we shall denote by
RK2,
h
(5) y m 1  y m  K 1  4K 2  K 3 
6
where K1  f (x m , y m )
h
h
K 2  f (x m  , y m  K1 )
2
2
K 3  f ( x m  h, y m  hK 1  2hK 2 )
which is three points Runge-Kutta
formula which we shall denote by
RK3.
In 1999 Goeken and Johnson[1]
introduced the following four points
Runge-Kutta method for finding the
numerical solutions of the autonomous
initial value problem of the ordinary
differential equation and it is of the
form
5
27
125
K3
(6) y m1  y m  K1  K 2
48
56
336
1
 K4)
24
where K1  hf ( y m )
1
1
K 2  hf ( y m  K 1  hf y K 1 )
3
18
152
252
K 3  hf ( y m 
K1 
K2
125
125
44

hf y K 1 )
125
19
22
25
K1  K 2  K 3
2
7
14
5
 hf y K 1 )
2
which we shall denote by GJ.
K 4  hf ( y m 
2
In 2003 Wu[2] introduced the two-step
formula Runge-Kutta method for
finding the numerical solution of the
autonomous initial value problem of
the ordinary differential equations and
it is of the form
3
1
1
(7) y m1  y m  y m1  h ( f m
2
2
2
121
8
23

f ( y m , hf m ) 
f (ym
192
11
192
44
44
8
 hf m  hf ( y m  hf m ))
23
23
11
121
20

f ( y m 1 
hf m 1 )
192
29
23
44

hf ( y m1  hf m1
192
23
44
8
 f ( y m 1  hf m 1 )))
23
11
which we shall denote by WU.
2
Problem Formulation
We know that the above equations (1)(2) are equivalent to the integral
equation
x
(8)
y( x )  c   f ( t, y)dt .
a
By the method of successive
approximation, the sequence y n (x)
converges to the solution of the above
differential equation where
x
(9)
y m1 ( x )  c   f ( t, y m ( t ))dt
a
So we will use the fact that the
sequence y n (x) converges to the
solution of the above differential
equation to approximate the numerical
solution at the point x  a  h by using
the numerical approximation of
a h
(10)
 f (t, y(t ))dt .
a
The four numerical integration
formulas that we will use in this paper
are;
b
ab
(11)
a y(x)dx  (b  a) y 2 
which is Midpoint formula,
b
1
(12)  y( x )dx  (b  a )y(a )  y(b)
2
a
which is Trapezoidal formula,
b
1
(13)  y( x )dx  (b  a )y(a )  y(b)
2
a
1
 (b  a ) 2 y(a )  y(b)
12
which is the Modified Trapezoidal
formula and
b
h
ab
(14)  y( x )dx   y(a )  4 y(
)
6
2
a
 y(b)
which is Simpson’s formula.
By equation (11), (12), (13) and (14),
we obtain four formulas which are
h
(15) y m 1  y m  hf ( x m  , y m  bb))
2
h
where bb  f ( x m , y m ) which we
2
shall denote by PS1,
h
y m 1  y m  K 1  K 2 
(16)
2
where K1  f (x m , y m )
K 2  hf (x m  h, y m  hf (x m , y m ))
which we shall denote by PS2 ,
h
y m 1  y m  K 1  K 2 
(17)
2
h2
 AA  BB
12
where K1  f (x m , y m )
K 2  hf (x m  h, y m  hf (x m , y m ))
AA  f x ( x m , y m )  f ( x m , y m )f y ( x m , y m )
BB  f x ( x m  h, y m  hf ( x m , y m ))  f ( x m  h,
y m  hf ( x m , y m ))f y ( x m  h, y m  hf ( x m , y m ))
which we shall denote by PS3 and
h
y m 1  y m  f ( x m , y m )
(18)
6
h
h
 4f ( x m  , y m  f ( x m , y m ))
2
2
 f (x m  h, y m  hf (x m , y m ))
which we shall denote by PS4.
3
3
Examples
There will be five examples in this
section. We will use the above nine
formulas to find the numerical
solutions of these five examples.
3.1
Example 1
Find the numerical solution of the
equation
4
(19)
y 
, x  [5,100]
x ( x  4)
with the initial condition
y(5)  5.
(20)
The analytical solution of the above
equation is
y( x )  5  ln( 5)  ln( x )  ln( x  4) .
The numerical results are in following
table 1, 2 and 3.
x  5.1, h  0.1
y  5.0755075525
RK1
RK2
RK3
PS1
PS2
@
PS3
PS4
RK1
RK2
RK3
PS1
PS2
PS3
PS4
Calculated y
5.0800000000
5.0756506239
5.0756507618
5.0754361150
5.0756506230
5.0756444162
5.0755076180
Table 1
x  5.1
h  0.0001
y  5.0755075525
Calculated y
5.0755119022
5.0755075527
5.0755076180
5.0755075525
5.0755075527
5.0755075527
5.0755075525
Table 2
Error
4.492447  10 3
1.430714  10 4
6.547634  10 8
7.143748  10 5
1.430714  10 4
1.368637  10 4
6.547634  10 8
Error
4.349691  10 6
1.891749  10 10
6.547634  10 8
5.820766  10 11
1.891749  10 10
1.891749  10 10
7.275958  10 12
RK1
RK2
RK3
PS1
PS2
PS3
PS4
x  100.0
h  0.0001
y  6.5686159179
Calculated y
Error
6.5686559406
4.002266  10 5
6.5686159595
4.156027  10 8
6.5686159586
4.071626  10 8
6.5686159165
1.462467  10 9
6.5686159595
4.156027  10 8
4.156027  10 8
6.5686159595
6.5686159586
4.071626  10 8
Table 3
3.2
Example 2
Find the numerical solution of the
equation
xy 3

y 
, x  [0,1.1]
(21)
1 x2
with the initial condition
y(0)  1.
(22)
The analytical solution of the above
1
equation is y( x ) 
.
2
3  2 1 x
The numerical results are in following
table 4, 5 and 6.
x  0.1 , h  0.1
y  1.0050251887
Eerror
Calculated y
RK1 1.0000000000
5.025189  10 3
RK2 1.0049751860
5.000273  10 5
RK3 1.0050377575
1.256881  10 5
PS1 1.0049937617
3.142699  10 5
PS2 1.0049751860
5.000273  10 5
PS3 1.0049627790
6.240965  10 5
PS4 1.0049875698
3.761890  10 5
Table 4
4
RK1
RK2
RK3
PS1
PS2
PS3
PS4
x  0.1
h  0.0001
y  1.0050251887
Calculated y
1.0050200075
1.0050251887
1.0050251887
1.0050251877
1.0050251887
1.0050251887
1.0050251887
Table 5
Eerror
5.101223  10 6
3.637979  10 12
3.637979  10 12
1.818989  10 11
3.637979  10 12
3.637979  10 12
1.818989  10 12
RK1
RK2
RK3
PS1
PS2
PS3
PS4
x  1.1
h  0.0001
y  6.1100397659
RK1
RK2
RK3
PS1
PS2
PS3
PS4
Calculated y
6.0616904647
6.1100181633
6.1100398300
6.1100009806
6.1100181633
6.1099903436
6.1100066900
Table 6
Eerror
4.834930  10 2
2.160255  10 5
6.413757  10 5
3.878533  10 5
2.160255  10 5
4.942231  10 5
3.307559  10 5
3.3
Example 3
Find the numerical solution of the
equation
sin x 2 y
y  2 
, x  [2,10]
(23)
x
x
with the initial condition
y(2)  1.
(24)
The analytical solution of the above
4  cos 2  cos x
equation is y( x ) 
.
x2
The numerical results are in following
table 7, 8 and 9.
RK1
RK2
RK3
PS1
PS2
PS3
PS4
x  2.1 , h  0.1
y  0.9271426912
Calculated y
0.9227324357
0.9272135323
0.9271409569
0.9273232808
0.9272135323
0.9273578742
0.9272866980
Table 7
x  2.1
h  0.0001
y  0.9271426912
Calculated y
0.9271386264
0.9271426912
0.9271426912
0.9271426913
0.9271426912
0.9271426912
0.9271426913
Table 8
Eerror
4.410255  10 3
7.084114  10 5
1.256881  10 6
1.805896  10 4
7.084114  10 5
2.151830  10 4
1.440068  10 4
Eerror
4.064775  10 6
6.002665  10 11
2.728484  10 12
1.673470  10 10
6.002665  10 11
6.002665  10 11
1.300577  10 10
x  10.0
h  0.0001
y  0.0442292469
RK1
RK2
RK3
PS1
PS2
PS3
PS4
Calculated y
0.0442252473
0.0442292467
0.0442294669
0.0442292468
0.0442292467
0.0442292467
0.0442292467
Table 9
Eerror
3.999578  10 6
1.901412  10 10
2.329243  10 10
1.350600  10 10
1.901412  10 10
1.901412  10 10
1.527951  10 10
3.4
Example 4
Find the numerical solution of the
equation
xy
y  2
, x  [0,100]
(25)
y  x2
with the initial condition
y ( 0)  1 .
(26)
5
The analytical solution of the above
equation is y(x)  x 2  x 4  1 .
The numerical results are in following
table 10, 11 and 12.
x  0.1 , h  0.1
y  1.0050124371
Calculated y
RK1 1.1000000000
RK2 1.0050505051
RK3 1.0050081470
PS1 1.0050125313
PS2 1.0050505051
PS3 1.0049572340
PS4 1.0050251872
Table 10
RK1
RK2
RK3
PS1
PS2
PS3
PS4
x  0.1
h  0.0001
1.0050124371
Calculated y
1.0050074292
1.0050124371
1.0050124371
1.0050124371
1.0050124371
1.0050124371
1.0050124371
Table 11
Eerror
9.498756  10 2
3.806794  10 5
4.290066  10 6
9.421456  10 8
7.084114  10 5
5.520314  10 5
1.275212  10 5
Eerror
5.007957  10 6
2.910383  10 11
9.094947  10 12
9.094947  10 12
2.910383  10 11
2.910383  10 11
1.818989  10 11
x  10.0
h  0.0001
y  0.0442292469
RK1
RK2
RK3
PS1
PS2
PS3
PS4
Calculated y
1414.2151819
1414.2151819
1414.2151819
1414.2151819
1414.2151819
1414.2151819
1414.2148829
Table 12
Eerror
1.619565  10 3
1.619574  10 3
1.619574  10 3
1.619574  10 3
1.619574  10 3
1.619574  10 3
1.318799  10 4
3.5
Example 5
Find the numerical solution of the
equation
(27)
y  y 3 , x  [0,49.5]
with the initial condition
y(0)  0.1
(28)
The analytical solution of the above
1
equation is y( x ) 
.
100  2x
The numerical results are in following
table 13, 14 and 15.
RK1
RK2
RK3
GJ
WU
*
PS1
PS2
PS3
PS4
RK1
RK2
RK3
GJ
WU*
PS1
PS2
PS3
PS4
x  0.1, h  0.1
y  0.10010015025
Calculated y
Eerror
0.10010000000
1.502503  10 7
0.10010015015
1.002718  10 10
0.10010015025
1.136868  10 13
0.10010030624
1.559881 10 7
0.10010584213
5.691884 10 6
0.10010015008
1.753051  10 10
0.10010015015
1.002718  10 10
0.10010015002
2.255547  10 10
0.10010015010
1.502940  10 10
Table 13
x  0.1
h  0.0001
y  0.10010015025
Calculated y
Eerror
0.10010015010 1.504077  10 10
0.10010025025 4.547474  10 13
0.10010025025 4.547474  10 13
0.10010025072 1.004693 10 7
0.10000010599 1.000443  10 4
0.10010025025 4.547474  10 13
0.10010025025 4.547474  10 13
0.10010025025 4.547474  10 13
0.10010025025 4.547474  10 13
Table 14
6
x  49.5
h  0.0001
y  1.0
Calculated y
Error
RK1 0.9996548547
3.451453  10 4
RK2 1.0000000038
3.845344  10 9
RK3 1.0000000088
8.847564  10 9
GJ
1.0003861906
3.861906  10 4
WU 17.085971673
16.085971673
*
PS1
0.9999999987
1.309672  10 10
PS2 1.0000000038
3.845344  10 9
PS3 1.0000000019
1.939043  10 9
PS4 1.0000000013
1.329681  10 9
Table 15
* WU formula used y(0) and y(0.5h )
as the initial two points.
4
Conclusion
All four new formulas are good
numerical formulas for finding the
numerical solutions of the initial
value problem of the ordinary
differential equations. These four
new formulas will give us more
freedom to find the way of finding
the numerical solutions of the initial
value problem of the ordinary
differential equations. However we
have to keep in mind that the above
numerical solutions are just for only
these five equations and the Wu
formula is the 2-step formula. We
strongly recommend the formula PS1
and the formula PS4.
References:
[1] David Goeken and Olin Johnson,
Fifth-Order Runge-Kutta with Higher
Order Derivative Approximations,
Electronic Journal of Differential
Equations, Vol.2, November 1999,
pp.1-9.
[2] Xinyuan Wu, A Class of RungeKutta of Order Three and Four with
Reduced Evaluations of Function,
Applied
Mathematics
and
Computation, Vol.146, December
2003, pp.417-432.