SOLVING INITIAL VALUE PROBLEM USING RUNGE-KUTTA
6
thORDER METHOD
Abbas Fadhil Abbas Al-Shimmary
Department of Electrical Engineering, College of Engineering, University of Al-Mustansiriyah, Bagdad, Iraq E-Mail: [email protected]
ABSTRACT
The initial value problems (IVPs) in ordinary differential equations are numerically solved by one step explicit methods for different order , the behavior of Runge- Kutta of third order method is obtained by Heun [4], Kutta found the complete classification and derivation of fourth and fifth order methods in [6], the derivation of sixth order method was found by Huta in [5] but in eight stages, Butcher in ([1], [2], [3]) presented the relation between the order conditions and the rooted trees up to sixth order. the main aim of this paper is to exhibit a new more simplest representation for the trees and the derivation of Runge- Kutta method of order six with seven stages including ( rooted trees, order condition and stability region), symbolically computations are used in the study to simplify the method, finally example illustrate the method are presented.
Keywords: initial value problems, ordinary differential equation, runge-kutta methods, order of methods.
1. INTRODUCTION
Ordinary differential equations arise frequently in several models of mathematical physics, biological sciences, engineering and applied mathematics. Unfortunately many cannot be solved exactly. This is why numerical treatment is very important and provides a powerful alternative tool for solving the differential equations. Which are modeled as initial value problems (IVPs)?There are various classes of methods in this direction. Taylor series method has a major disadvantage ,it requires evaluation of partial derivatives of higher orders manually and this is not possible in any practical application, therefore we seek for improved methods which do not need evaluation for repeated differentiation of the differential equation, Although the famous paper by Runge, and subsequently developed by Heun and Kutta , still the explicit Runge-Kutta of the 4th order method have
been widely used and the most popular version is the classical 4th order, the Runge paper is now recognized as
the starting point for modern one-step methods with multivalued and multistage, construction of this method needs the derivation and solution of many nonlinear algebraic system of equations called the order conditions , thus the calculation process were performed using Mathematica program, this system of nonlinear algebraic equations has many solutions, therefore we must guess the integration step size using trial and error to estimate the truncation error ,this is the main drawback of Runge-Kutta methods. a simplified derivation of fourth order Runge-Kutta method given in [14], the derivation of fifth order method were introduce by Kutta [4] and corrected by Nystrom [6], and the sixth order with eight stages founded by Huta [3], this is a short brief history of the method also we find that it is impossible to present a general formula to the order conditions for all families of Runge-Kutta methods but there is a connection can be presented in this paper between the several sequences orders.
derivation of the explicit Runge - Kutta - Butcher of the six order method, in sect. 3 we present the simplest new tree representation related to differential elementary for the order conditions of sixth orders with seven stages. sect. 4 exhibits the linear combination between the order conditions. The stability region of the method is presented in sec. 5; finally we give numerical test example in sect. 6.
2. OUTLINE AND MATHEMATICAL DERIVATION OF THE METHOD
In this section we will introduce the basic tools required the derivation to obtain the order condition for Runge -Kutta of six orders with seven stages method, the typically general form of the scalar first order of ordinary differential equation is
))
(
,
(
x
y
x
f
y
,y
(
x
0)
y
0(1)
Which is known with initial condition ”the initial value problem (IVB)”,the main aim in this section is to obtain approximate solution of
y
(
x
)
, the approach of the sixth order with seven stages Runge-Kutta method is proceed to evaluatey
n1 as an approximation to)
(
)
(
x
1y
x
h
y
n
n
, an important special case “ without loss of generality” if the function f does notdepend on x but only of y ,by setting
x
1
then equation (1) reduces to so called “autonomous” , and written in the form))
(
(
)
(
x
f
y
x
y
,y
(
x
0)
y
0 (2)
7 1 0 1 i i iK
b
h
y
y
(3)and the exact solution of equation (2) has the form
7 1 20
)
O(h
)
(
j ij i
y
x
h
a
Y
(4)
and according to Taylor series expansion , equation (4) take the form
7 1 0)
(
i i ij iy
x
h
a
K
Y
(5)
where Ki are slops define as below , for finding y1required
the next seven intermediate calculations:
)
(
)
)
(
,
(
)
)
(
,
(
)
)
(
,
(
(6)
)
)
(
,
(
)
)
(
,
(
)
,
(
)
,
(
7 7 6 6 5 5 4 4 3 3 2 2 1 1 1 6 76 5 75 4 74 3 73 2 72 1 71 0 7 0 7 5 65 4 64 3 63 2 62 1 61 0 6 0 6 4 54 3 53 2 52 1 51 0 5 0 5 3 43 2 42 1 41 0 4 0 4 2 32 1 31 0 3 0 3 1 21 0 2 0 2 0 0 1
y
h
b
K
b
K
b
K
b
K
b
K
b
K
b
K
y
K
a
K
a
K
a
K
a
K
a
K
a
h
y
h
c
x
f
K
K
a
K
a
K
a
K
a
K
a
h
y
h
c
x
f
K
K
a
K
a
K
a
K
a
h
y
h
c
x
f
K
K
a
K
a
K
a
h
y
h
c
x
f
K
K
a
K
a
h
y
h
c
x
f
K
K
ha
y
h
c
x
f
K
y
x
f
K
n nWhere the vector C indicated to the nodes
weights within the step (typically
c
j
[
0
,
1
]
) ,the matrixA indicated to the dependence of the stages on derivatives found at previous stages ( and the row sum condition hold
2,3,...,7
j
,
)
(
1 1
j k j jkc
a
), the vector BT indicated tothe quadrature weights(
7 11
i ib
),which the final resultdepend on the derivatives, the approach of RK methods is expanded equation (5) in Taylor’s series expansion , after some algebraic simplification this expansion is equated to the exact solution y(x0h)that is also given by the Taylor’s series ) ( ) ( ! 6 1 ) ( ! 2 1 ) ( ) ( ) ( 7 0 ) 6 ( 6 0 2 0 0 0 h O x y h x y h x y h x y h x y (7)
the first step is to calculate the successive derivatives of
,
,
y
y
up to order six using chain rule for equation(2) as the following:)
(
))
(
(
)
(
x
f
y
x
y
x
f
f
y
y(8)
)
(
x
f
f
f
f
yf
f
y
yy
y(9) 3 ) ( ) 4 ( f f f f ff f f f f f fff f x
y yyy yy y y yy y y y (10)
(11) 3 4 6 4 3 ) ( ) 5 ( f f f f f ff f f f ff f f f ff f f f fff f f fff f f f ff f f fff f f ffff f x
y yyyy y yyy yy y y yy yy yyy y yy y y y yy y y y yy y y y y
(12) 2 2 3 3 4 4 7 2 4 3 3 4 4 7 5 7 7 8 3 7 ) ( ) 6 ( f f f f f f ff f f f f fff f f f ff ff f f ffff f f ff f f f f f ff f f f ff ff f f f f ff f f f fff f f ff f ff f f f ff f f f f f ff f ff f ff f ff ff f f fff ff f f f f f ff f ff f ff f fff f ff f f fff f f f f fff f ff fff f f ff ff f ff fff f ffff f f fffff f x y y y y y y yy y y y yyyy y y y yyy y yyyy y y yy y y y y yy y yy yy y y y yy y y yyy y y yy yy y y y yy y y y yy yy y yy yy y yy yyy y yy y yy y yy y y yyy yy y yyy y y yyy y y yyy yy yyy y y yyy y yyyyy y yyyy yyyyyy
(13) ) ( ... )
(
) (
2 1
) ( ) (
7 3
7
3 2
1 7
2
3 2
7
2 2 7
1 0 1
h O f
f f h c a b
f f f h c b
f f h c b hf
b y
y
y y i ji
j ij i
yy i
i
y i
i i i
i
i
since equation (1) is scalar autonomous problem , so we may use a simplified approach for instance
2
f
f
f
ff
f
f
ff
f
f
y yy
yy y
y yy(see [4]), expressions
like
f
,
f
yf
,
f
yy(
f
,
f
)
,
f
yf
yf
,...
are known as “elementary differentials” they are related to a rooted trees, which is defined in the next section.3. ROOTED TREE THEORY
Firstly we will classify the nodes for trees into a new three representation types, the first kind called the initial node (root) which is usually exist in each tree and denoted it by the symbol, the second type is called the intermediate nodes (non-leaf nodes that take the symbol), the third type of nodes is named the leafs (terminal nodes that take the symbol). In this section we need to give some important definitions” A rooted tree is a set of p number of the different types of nodes (pequal to the order of the tree is the measure of how much nodes the tree has) connected by edges oriented away from the specific node called (root)”, so it is a new simplest combinatorial graphs that contain no cycles (no connection between initial and terminal nodes). Graphically we represent a tree that
corresponding to an elementary differential, for a root
node labelled by
f
yyy...y (we mean the kth partialderivatives of the function f w.r.t independent variable y) attached to k sub trees corresponding to those sub lists, lists with no sub lists are leaves. Each tree is named beginning from the initial node (root) passing through the intermediate nodes (or brunches ) to the terminal nodes (leaves) and the index start from down node (root) successively to upper nodes depending on the number of branches that growths except the leaves nodes and from left branch to right, for this purpose we present in (figure 1) below a simple example of a treet of order 7 (number of total different types of nodes is 7), always the root node labelled by
i
, the non-leaf nodes labelled by (j
,
k
,
and
l
) and neglect the leaves without label,
)
(
t
F
is the elementary differential function given in terms of an expression of seven components (number of components is equal to the order tree),
(
t
)
is the weighted function that correspond to each elementary differential formulaF
(
t
)
and
(
t
)
is the density of the tree which equal to the product of the numbers nodesabove.
T t
i
i
t
t) ( )
(
where
(
t
i)
are the density ofthe sub trees above, the symmetry of the tree
(
t
)
is the set of permutation generated by all members of symmetry for the tree t (the mapping
(
t
)
:
T
T
that preserves the root and the structure ) which equal to how the tree symmetric hasFigure-1. Sample tree of order seven with (elementary differential and weight) functions and its density.
below we illustrate the correspondence between elementary differential and rooted trees using both formulation in a Table for trees up to order six with its density.
Table-1. Illustrate the relation between elementary differential and rooted trees up to order 6.
Order 1 2 3 3 4 4 4 4 5
F(t) F
f
yf
f
yyff
f
yf
yf
f
yyyfff
f
yyff
yf
f
yf
yyff
f
yf
yf
yf
f
yyyyffff
Order 5 5 5 5 5 5 5
F(t)
f
yyyfff
yf
f
yyf
yff
yf
f
yyff
yyff
f
yyff
yf
yf
f
yf
yf
yyff
f
yf
yyyfff
f
yf
yyff
yf
tree
)
(
t
10 20 15 30 60 20 40Order 5 6 6 6 6 6 6
F(t)
f
yf
yf
yf
yf
f
yyyyyfffff
f
ffff
f
yyyy yf
ff
ff
f
yyy y yf
yyyfff
yf
yf
f
yf
yf
yf
yyff
f
yf
yf
yyyfff
Tree
)
(
t
120 6 12 24 12 360 120Order 6 6 6 6 6 6 6
F(t)
f
yf
yf
yyff
yf
f
yyf
yff
yyff
ffff
f
f
y yyyyf
yyf
yff
yf
yf
ff
f
ff
f
yy y yyf
yyff
yyyfff
f
yyff
yyff
yf
Tree
)
(
t
240 36 30 72 72 24 48Order 6 6 6 6 6 6 6
F(t)
f
yyyfff
yyff
f
f
f
ff
f
y y yy yf
f
ff
f
f
y yy y yf
yyff
yf
yf
yf
f
f
fff
f
y yyy yff
ff
f
f
y yy yyf
yf
yf
yf
yf
yf
Tree
)
(
t
18 120 180 144 60 90 720We can write equations (3) and (5) in terms of the
weighted functions as follows:
(
)(
)
(13)
!
)
(
)
(
)
(
)
(
6 01
) (
0
0
F
t
y
p
t
h
t
x
y
h
x
y
p
p
(14)
)
)(
(
!
)
(
)
(
0 6
1
) (
0
1
F
t
y
p
h
t
t
y
y
p
p
Where
(15)
(t)
p!
(t)
and
)
(
)
(
!
)
(
t
t
p
t
a Runge-Kutta of order sixth with seven stages are
converges if
)
(
1
)
(
t
t
by comparing equation (13) with (14) , see [2].the results are a system of nonlinear algebraic equations which is denoted by the order conditions ,which is arranged in this
paper in regular arrangement and sequential up to sixth order with seven stages , the following twenty two conditions must be satisfied equations(16a,…,16e) in addition to the constraints given by equation (15)
(16a)
6
1
,
3
1
5
1
c
b
,
2
1
c
b
4
1
,
1
5 7
2 2
7
2
4 7
2 i
i i
7
2 i
i
3 7
2 7
1
i
i i
i
c
b
c
b
c
b
b
i i i
i
i i i
i
(16b)
)
16
(
120
1
!
6
!
3
]
)
(
[
60
1
!
5
!
2
]
)
(
[
24
1
!
4
!
1
]
)
(
[
7
4 2
2
3 1
1 7
4 2
2
2 1
1 7
4 2
2 1
1
c
c
a
a
b
c
a
a
b
c
a
a
b
i k i k
i j
jk ij i
i k i k
i j
jk ij i
i k i
k k
i j
jk ij i
k k
)
16
(
360
1
!
6
!
2
]
)
(
[
120
1
!
5
!
1
]
)
(
[
7
5 2
3
2 1
2 1
1 7
5 2
3 1
2 1
1
d
c
a
a
a
b
c
a
a
a
b
i l i l
i k
kl k
i j
jk ij i
i l i
l l
i k
kl k
i j
jk ij i
l
)
16
(
720
1
!
6
!
1
]
)
(
[
7
6 2
4 1
3 1
2 1
1
e
c
a
a
a
a
b
i mi
m m
i l
l
i k
lm kl k
i j
jk ij
i
and solve many of them that satisfy the sixth order Runge-Kutta method and this process calculation required using Mathematica program.
4. REDUCTION THE ORDER CONDITIONS
Since the number of rooted trees being increase for order greater than four, the number of conditions increase and complexity for order six, for this reason it is necessary to find a relationships (linear combination) between the results order conditions corresponding to different trees, for example the two particular trees t1 and
t2 with the same six order, which differ only in small
respect
Figure-2. Illustrate two equivalent particular trees. Hence
(
t
1)
,
(
t
2)
are equivalent if
22
1
i
c
c
a
ij j and the linear combination between thetwo conditions
2
1
b
ic
j and2
1
2
b
ic
j givenby
6
1
)
1
(
b
i
c
ic
i
, this result state that two trees are equivalent if
(
t
1)
that corresponds t1 can be expressed in terms of
(
t
2)
that corresponds t2and trees
of
lower order. Similarly we listed all conditions which can be written as a linear combination and satisfy the Runge-Kutta method of order six and seven stages below.We assume the modified conditions C(2),D(1)
...(17)
1,2
and
1,2,...,7
for
c
1
)
2
(
71 1
l
i
k
c
a
C
li i
l j ij
(18)
1,2,...,7
for
)
-(1
b
)
1
(
7b
a
c
j
applied to all stages ,the only condition that remain
)
(
1
)
(
t
t
for the treesFigure-3. Illustrate the trees that cannot be written as linear combinations.
The linear combinations appear in equations (19e-19h) below; we list all conditions which are needed to specify the method.
(19a)
...
0
b
,
6
1
)
1
(
,
2
1
,
1
2 7
1
7
1 7
1
i
i i i
i i i i
i
c
c
b
c
b
b
(19b)
6
12
20
30
1
)
)(
)(
)(
1
(
6 4 3 6 4 6 3 4 3 6 4 3
4 3
7
1
6
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
b
i i ii
i i i
9c)
...(1
6
12
20
1
)
(
)
)(
1
(
6 4 6 4
4 7
1
6
c
c
c
c
c
c
c
c
c
c
b
i ii
i i i
d)
...(19
6
12
1
)
)(
1
(
67
1 6
c
c
c
c
c
b
ii
i i
i
...(19e)
24
60
1
)
(
)
1
(
33 7
1 ,
c
c
c
c
a
c
b
j jj i
ij i
i
...(19f)
24
60
40
90
1
)
(
)
)(
1
(
6 3 6 3
3 7
1 ,
6
c
c
c
c
c
c
c
a
c
c
c
b
j jj i
ij i
i i
)
...(19g
24
60
120
1
)
)(
(
)
1
(
4 3 4 3
3 7
1 ,
4
c
c
c
c
c
c
c
c
c
a
c
b
j jj i
j ij i i
)
...(19h
120
360
1
)
(
)
1
(
3 3 7
1 , ,
c
c
c
c
a
a
c
b
k kk j i
jk ij i i
)
...(19i
2
i
,
2
1
27
1
j i
j
ij
c
c
a
19j)
...(
7
1 ,
i j
i
ij
c
a
9k)
...(1
)
1
(
71 ,
j i j
i ij
i
a
b
c
b
L)
...(19
0
)
1
(
7
1 2
i
i i i
c
a
b
9m)
...(1
0
)
1
(
7
1
2
i
i i i i
c
c
a
b
19n)
...(
0
)
1
(
71 ,
2
j i
i ij i i
c
a
a
b
We can solve for the components bT from
equations (19a-19d) ,then solve for a54 ,a64 and a65 from
the consistent equations (19e-19h), we then solve for a32
from (19i) and then for a42 ,a43,a52,a53,a62 and a63 from
200
13
25
4
25
4
40
11
40
11
0
200
13
39
80
195
32
39
118
156
43
13
33
260
261
1
10
1
2
1
8
1
20
11
20
3
6
1
8
15
48
35
24
55
48
25
6
5
12
1
3
1
12
1
3
1
3
2
0
3
2
3
1
3
1
0
5. STABILITY REGIONIn this section we will study in addition to the numerical method converges to the exact solution over a bounded interval, the qualitative or stability analysis that assign to the fitness of the method with the used numerical solution, because the complexity of this type of analysis we restrike to the linear problem with constant coefficients so called the “stiff problem” or test equation where
is a constant and also complex in nature.n
n
y
y
f
(
)
K
1
(21))
3
/
1
(
)
3
/
(
K
2
f
y
n
h
K
1
y
n
h
(22)))
3
/
1
(
3
/
2
1
(
)
3
/
2
(
K
3 2
y
h
h
K
h
y
f
n n
(23))))
3
/
1
(
3
/
2
1
(
12
/
)
3
/
1
(
3
/
12
/
1
(
)
12
/
3
/
12
/
(
K
4 1 2 3
y
h
h
h
h
h
h
K
h
K
h
K
h
y
f
n n
(24)))))
3
/
1
(
3
/
2
1
(
12
/
)
3
/
1
(
3
/
12
/
1
(
8
/
15
))
3
/
1
(
3
/
2
1
(
48
/
35
)
3
/
1
(
24
/
55
48
/
25
1
(
)
8
/
15
48
/
35
24
/
55
48
/
25
(
K
5 1 2 3 4
h
h
h
h
h
h
h
h
h
h
h
h
h
y
K
h
K
h
K
h
K
h
y
f
n n
(25))))))
3
/
1
(
3
/
2
1
(
12
/
)
3
/
1
(
3
/
12
/
1
(
8
/
15
))
3
/
1
(
3
/
2
1
(
48
/
35
)
3
/
1
(
24
/
55
48
/
25
1
(
10
/
)))
3
/
1
(
3
/
2
1
(
12
/
)
3
/
1
(
3
/
12
/
1
(
2
/
))
3
/
1
(
3
/
2
1
(
8
/
)
3
/
1
(
20
/
11
20
/
3
1
(
)
10
/
2
/
8
/
20
/
11
20
/
3
(
K
6 1 2 3 4 5
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
y
K
h
K
h
K
h
K
h
K
h
y
f
n n
(26) )))))) 3 / 1 ( 3 / 2 1 ( 12 / ) 3 / 1 ( 3 / 12 / 1 ( 8 / 15 )) 3 / 1 ( 3 / 2 1 ( 48 / 35 ) 3 / 1 ( 24 / 55 48 / 25 1 ( 10 / ))) 3 / 1 ( 3 / 2 1 ( 12 / ) 3 / 1 ( 3 / 12 / 1 ( 2 / )) 3 / 1 ( 3 / 2 1 ( 8 / ) 3 / 1 ( 20 / 11 20 / 3 391 / 80 (27) )))) 3 / 1 ( 3 / 2 1 ( 12 / ) 3 / 1 ( 3 / 12 / 1 ( 8 / 15 )) 3 / 1 ( 3 / 2 1 ( 48 / 35 ) 3 / 1 ( 24 / 55 48 / 25 1 ( 95 1 / 32 ))) 3 / 1 ( 3 / 2 1 ( 12 / ) 3 / 1 ( 3 / 12 / 1 ( 39 / 118 )) 3 / 1 ( 3 / 2 1 ( 15 / 43 ) 3 / 1 ( 13 / 33 260 / 261 1 ( ) 39 / 80 195 / 32 39 / 118 156 / 43 13 / 33 260 / 261 (K7 1 2 3 4 5 6
h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h y K h K h K h K h K h K h y f n n we get
)
54
5
18
5
3
5
27
203
45
1346
15
1456
200
(
200
/
y
2 3 4 5 61
n
y
n
h
z
z
z
z
z
z
(29)
simplified and divide both side by
y
n then the stability polynomialQ
(z)
y
n1/
y
ngiven by
) 2160
1 720
1 120
1 5400
203 4500
673 375 182 1 )
( 2 3 4 5 6 7
z z z z z z z z
Q
(30)
Figure-4 show the stability region of the six order seven stages of Runge- Kutta methodin this stability region,the range for the real part of
is
3
.
5
Re(
z
)
0
.
0
.Figure-4. the stability region of the Runge- Kutta method of six order seven stages.
5. NUMERICAL PROBLEM
In this section, we tasted the numerical solution of initial value problem where the method of Runge-Kutta of order six with seven stages is applied to show the efficiency and the approximate of the method and compared the result with the exact solution. The problem are solved for
x
[
0
.
1
,
0
.
5
]
Problem 5.1 Solve the initial value problem
1
y(0)
,
sin
y
y
,using Runge - Kutta method oforder six with seven stages, where the exact solution given by
604582
.
0
coty)
-ln(cscy
x(y)
see (31)Table-2. Illustrate the numerical solution of example 5.1 using RK 6th order with seven stages.
X yusing order six yexact Error
0.1 1.08636 1.0863565 ‐0.000004 0.2 1.17682 1.1768209 0.000001 0.3 1.27082 1.270817 ‐0.000003 0.4 1.36763 1.367627 ‐0.000003 0.5 1.46641 1.466404 ‐0.000006
6. CONCLUSIONS
In this paper we have simplified the trees representation which related to elementary differentials that it uses in derivation Runge-Kutta methods and to reduce the complexity that arise by indexing the trees, it should be mentioned that the classical Runge-Kutta process of order four may be still more economical by the derivation than the six order.
ACKNOWLEDGEMENTS
The author would like to thank Al-Mustansiriyah University (www.uomustansiriyah.edu.iq) Baghdad- Iraq for its support in the present work.
REFERENCES
[1] J. C. Butcher. 2008. Numerical Methods for Ordinary Differential Equations, 2ndEd.Wiley.
[2] J. C. Butcher. 2009. On fifth and Sixth OrderExplicit Runge-Kutta MethodsOrderCondition and Order Barriers. Canadian Applied Mathematics Quarterly. 17(3).
[3] J.C. Butcher. 2000. Numerical methods for ordinary differential equations in the 20th century. J. of
Comput. and Applied Math. 125: 1-29.
[4] K. Heun. 1900. Neue Methoden zur approximativen Integration der Differential- gleichungen einer unabängigen Veränderlichen, Z. Math. Phys. 45: 23-38.
[6] W. Kutta. 1901. Beitrag zur näherungsweisen Integration totaler Differential- gleichungen, Z. Math. Phys. 46: 435-453.
[7] Papageorgiou G. and Tsitouras Ch. 2002. Runge-Kutta Pairs for scalarautonomous nitial Value Problems Int. J. Comput. Math.80: 201-209.
[8] C. Runge. 1895. Über die numerische Auflösung von Differential gleichungen, Math. Ann.46: 167-178.
[9] E.J. Nyström. 1925. Über die numerische integration von differential gleichungen Acta. Soc. Sci. Fennicae. 50(13): 55.
[10]Daniel I. Okunbor,Robert D. Skeel. 1994. Canonical Runge-Kutta- Nyström methods of order five and six, j. of Comp. and Applied Math. 51: 375-382.
[11]Kalaf Bashir M.,Ismail Mohammed M. 2008. Runge-Kutta methods of high order for solving stiff problems. Al-Rafiden J. of computer sciences and Mathematics. p. 5.
[12]Mohammed M. Ismail. 2011. Goeken-Johnson six-order Runge-Kutta method, J. Edu. & Sci. p. 240.
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