INTRODUCTION
A class of twostep hybrid methods for the numerical integration of second order initial value problem
is defined in the following. Let h>0 denote the stepsize t_{n} = t_{0}+nh, h = 0,1,2,… and set y_{n} = y(t_{n}), f_{n} = f(t_{n},y_{n}).
We consider the method of the form
where:
The methods are fourth order accurate if satisfy the following conditions.
where:
and n_{14} = s_{+}+s+u_{+}+u_{–} and the
local truncation error is
where,
and
Further from condition (ii) either β_{1} = 0 or A_{+}+A_{–}C_{+}C_{–} = 0. Also from condition (vii) either β_{1} = 0 or α_{1} = 0 or A_{+}+A_{–}C_{+}C_{–} = 0. We consider the case where β_{1} = 0 and β_{1}≠0.
Case 1: If β_{1} = 0 then from condition (iii) β_{0} = 1/12 and from condition (i) γ = 5/6. Then the local truncation error from (6) becomes
Case 2: If β_{1}≠ 0 then either α_{1} =
0 or α_{1} ≠ 0.
(a) If α_{1} = 0 then we have from the conditions
and the LTE becomes
where,
and
Observe that Chawla and Rao^{[1]} method is of this class. They choose
the parameters as
and
LTE
(b) If α_{1}≠0 then fourth order conditions becomes
and the local truncation error becomes
where
PStability: Lambert and Watson^{[2]} have given the definition of Pstability
To analyses the stability properties of the method, we apply the method (2), (3), (4) to the scalar test equation:
We obtain the stability polynomial
where:
Equation (14) is a difference equation of the form^{[3]}
where:
When we apply the fourth order condition then the stability polynomial is of
the form
where,
Making the RouthHerwitz transformation
(15) becomes
Thus, the necessary and sufficient conditions for Pstability are
Since
then the condition becomes
and the necessary and sufficient conditions for Pstability are
and
where:
and
Condition (18) is satisfied iff
Thus the methods are Pstable if satisfy the condition (17) and (18) with
Phase LAG: When the method (2), (3) and (4) is applied to the scalar
equation (13), we have the recurrence relation^{[4]}:
Substituting we
have
on expansion of and , we have the form
Let
Substitutes in (19) and then comparing the coefficients of H^{j}, j
= 2, 3,4,5,6 to zero, we get
Thus we have
If
which is greater than 1/576 (Pstability condition), we may write
and on substituting in (19) and then comparing the coefficients of H^{7}
and H^{8} to zero, we get η_{5} = 0 and
Thus the quantity b1 in the definition of phase lag is
PARTICULAR METHODS
Case 1: β_{1} ≠ 0 and α_{1} = 0 Observe that Chawla and Rao^{[1]} method is of this class with parameters.
In this case
satisfy the Pstability condition and also phase Lag given by (20).
We choose the parameters in two different ways
(a) 
The points (t_{nα1}, y_{nα1})
and (t_{n+α1}, y_{n+α1}) are coincident and 
(b) 

(a) 
If the points (t_{nα1}, y_{nα1}) and (t_{n+α1},
y_{n+α1}) are coincident then 
then from (8)
for Pstability we have
and if we have
then the method has phase lag given by (20). Thus we take
for Pstable method. We choose β_{1} = 1, then s_{+} =
1/720. Also let A_{+} = 1/2.
(b) 
For
we have 
then from (8)
and for Pstable with phase lag given by (20), we must have .
Thus we choose β_{1} = 1, then s_{+} = 1/360. Also let
A_{+} = 1/2
Also we test the method with A_{+} = 0 and β_{1} = 1/3, then s_{+} = 1/120. The local truncation error for all these method is
Case 2: If β_{1} ≠ 0 and α_{1} ≠ 0,
we have chosen the method for α_{1} = 1. In this case Then
then from (11), we have
For Pstable method with phase lag of order eight given by (20), we must have
or in this case,
We choose β_{1} = 1 and A_{+} = 1/2, then,
For these methods, LTE is given by (26).
Methods for different examples were tested. The methods are given below.
The methods derived above and used for companion purpose in next section.
M 1: Given by equations (22) and (23)
with parameters β_{1} = 1, S_{+} = 1/720 and A_{+}
= 1/2
M 2: Given by equations (24) and (25)
with parameters β_{1} = 1, S_{+} = 1/360 and A_{+}
= 1/2
M 3: Given by equations (24) and (25)
with parameters β_{1} = 1/3, S_{+} = 1/120 and A_{+}
= 0
M 4: Given by equations (27) and (28)
with parameters β_{1} = 1, S_{+} = 1/360 and A_{+}
= 1/2
M 5: Given by equations (21).
NUMERICAL ILLUSTRATION
We have tried a number of explicit scalar (nonstiff) test problems of the form
(1). They give similar results and so we restrict our attention to one oscillatory
example.
Example 1: 

This is a pure oscillation problem whose solution has maximum amplitude unity
and period approximately six. We have calculated error as Error at t = 6
In Table 13 we present the following statistics:
1) Number of evaluation of the differential equation right hand side f, FCN;
2) Number of steps overall, NOST;
3) Number of successful steps to complete the integration, NSST;
4) Number of steps where the stepsize is changes, NCST;
5) Number of failed steps, NFST;
6) Number of steps on which the iteration diverged, NDIV;
7) Number of steps where the stepsize is halved, NHST;
Table 1: 
Comparison of the method for example 1 with Tol = 10^{2} 

Table 2: 
Comparison of the method for example 1 with Tol = 10^{4} 

Table 3: 
Comparison of the method for example 1 with with Tol = 10^{6} 

Next we consider example for system, where we use a test problem a moderately
stiff system of two equations.
Example 2: 

For this example we have deliberately introduced coupling from the stiff (linear)
equation to the nonstiff (nonlinear) equation. For this example again we have
calculated error as Error at t = 6_{∞}.
Table 4: 
Comparison of the method for example 2 with with Tol = 10^{2} 

Table 5: 
Comparison of the method for example 2 with with Tol = 10^{4} 

Table 6: 
Comparison of the method for example 2 with with Tol = 10^{6} 

Example 3: Consider the almost periodic linear problem studied by Stiefel
and Bettis^{[5]}.
with analytic solution.
Z(t) = cost+0.0005tsint+i(sin t0.0005t cost).
This solution represents motion on a perturbed circular orbit with the distance
from the origin Z(t). We have computed solution to this problem using our
fourth order method. We use the system
Where, Z = u+iv. After computing u and v, we have calculated Z at t = 40π
with stepsize h = π/4. Result are presented in Table 79
and comparison of approximation in Table 10.
Table 7: 
Comparison of the method for example 3 with with Tol = 10^{2} 

Table 8: 
Comparison of the method for example 3 with with Tol = 10^{4} 

Table 9: 
Comparison of the method for example 3 with with Tol = 10^{6} 

Table 10: 
Comparison of the approximation produced at Z=40π with
h=π/4 initially.Exact value is Z(40π)=1.0019720 
