ABSTRACT

In this communication we have used Bickley’s method for the construction of a sixth order spline function and apply it to solve the linear fifth order differential equations of the form where and are given functions with the two different problems of different boundary conditions. The method is illustrated by applying it to solve some problems to demonstrate the application of the methods discussed.

1. Introduction

In the recent past, several authors have considered the application of cubic spline functions for the solution of two point boundary value problems. Bickley [1] has considered the use of cubic spline for solving second order two point boundary value problems. The essential feature of his analysis is that it leads to the solution of a set of linear equations whose matrix coefficients are of upper Heisenberg form. Bickley uses a special notation other than the conventional one for the representation of the cubic spline, for a detailed discussion one may refer to E. A. Boquez and J. D. A. Walker [2], M. M. Chawla [3], and P. S. Ramachandra Rao [4-7]. We used Bickley’s method for the construction of a sixth degree spline and apply it to the linear fifth order differential equation with two different problems with different boundary conditions. The work has been illustrated through examples with h = 0.5 and h = 0.25.

2. Cubic Spline-Bickley’s Method

Suppose the interval is divided in to n subintervals with knots starting at, the function in the interval is represented by a cubic spline in the form

(1)

Proceeding in to the next interval, we add a term; proceeding in to the next interval, we add another term and so until we reach. Thus the function  is represented in the form for

     (1.1)   (1.2)         (1.3)

2.1. The Two-Point Second Order Boundary Value Problem

First, we consider the linear differential equation

      (1.4)

With the boundary conditions

         (1.5)

The number of coefficients in (1.1) is (n + 3). The satisfaction of the differential equation by the spline function at the (n + 1) nodes gives (n + 1) equations in the  (n + 3) unknowns. Also the end conditions (1.5) give us two more equations in the unknowns. Thus we get (n + 3) equations in (n + 3) unknowns. after determining these unknowns we substitute them in (1.1) and thus we get the cubic spline approximation of. Putting in the spline function thus determined, we get the solution at the nodes. The system of equations to be satisfied by the coefficients are derived below.

Substituting (1.1), (1.2), (1.3) in (1.4), at we get

                          (1.6)

where and so on. Applying boundary conditions in (1.5), we get

(1.7)

If these equations are taken in the order (1.7), (1.6) with, the matrix of the coefficients of the unknowns is of the Heisenberg form, namely an upper triangle with a single lower sub-diagonal. The forward elimination is then simple, with only one multiplier at each step and the back substitution is correspondingly easy.

2.2. Construction of the Sixth Degree Spline

Suppose the interval is divided in to “n” subintervals with knots. Starting at x₀, the function in the interval is represented by a sixth degree spline

Proceeding in to the next interval, we add a term, Proceeding in to the next interval we add another term and so until we reach. Thus the function is represented in the form

  (1.8)

It can be seen that and its first five derivatives are continuous across nodes.

3. Fifth Order Boundary Value Problem

We consider the linear fifth order differential equation

          (1.9)

With the boundary conditions

     (1.10)

We get (n + 6) equations in (n + 6) unknowns,. After determining these unknowns we substitute them in (1.8) and thus we get the sixth degree spline approximation of. Putting in the spline function thus determined, we get the solution at the nodes. The system of equations to be satisfied by the coefficients are derived below. From (1.8) we get

(1.11)

using (1.8) & (1.11) in the differential Equation (1.9) at the nodes takes of the form

(1.12)

To these equations we add those obtained from the boundary conditions (1.10), we get

                 (1.13) (1.14)                (1.15) (1.16)                 (1.17)

If these equations are taken in the order (1.14), (1.16), (1.12) with, (1.17), (1.15) & (1.13) the matrix of the coefficients of the unknowns, , is an upper triangular matrix with two lower sub diagonals. The forward elimination is then simple with only two multipliers at each step, and the back substitution is correspondingly easy.

3.1. Example 1

Consider the following fifth order linear boundary value problem

  (1.18)

With the boundary conditions

  (2)

by taking equal subintervals with h = 0.5 and h = 0.25 1) Solution with h = 0.5 The sixth order spline which approximates is given by

   (3)

where. We have eight unknowns and eight conditions to be satisfied by these unknowns are,

            (4)   (5)

Since it follows that Equation (3) reduces to the form

 (6)

also since and equations of (5) for i = 0 & 2 reduces to

and

It follows that we have to determine the five unknowns in Equation (6), subject to the five conditions

(7)

from (6)

(8)

and

  (9)

Substituting (6), (8), (9) in (7) we get the system of equations

(10)

Solving these we get

Substituting these values in (6) we get

(11)

where h = 0.5 Therefore

The analytical solution of the differential equation (1.18) subject to the conditions is given by

          (11.1)

The exact value of

It follows that the Absolute error of the numerical value of, computed from the spline approximation is 0.00083433 which is very small.

2) Solution with h = 0.25

The interval [0,1] is divided in to 4 equal subintervals we denote the knots by where,.

The sixth order spline which approximate is given by

  (12)

There are 10 unknowns in which are to be determined from 10 conditions

 (13)

In view of the conditions and it follows that hence The spline reduces to the form

(14)

From (14) 

(15)   (16)

Substituting (14), (15), (16) in (13) taken in the order,

we get the following system of equations

(17)

From the above system of equations, we notice that the coefficient matrix is an upper triangular matrix with two lower sub diagonals. solving the above equations we get

  (18)

However it may be noticed that from the Equation (17) which when substituted in the remaining equations will give us a 6 × 6 system of equations which may be solved. Substituting (18) in (14) we get the spline Approximation of. The values of, and The corresponding absolute errors at tabulated in Table 1.

The analytical solution of the differential equation (1.18) with the conditions is given by (11.1) is symmetric about the central value. The same aspect is also satisfied by the numerical approximations as is evident from the above table. We found that the approximate values are remarkably accurate.

3.2. Example 2

Consider the following fifth order linear boundary value problem

   (19)

Subject to

   (20)

1) Solution with

The sixth order spline which approximates is given by (3). The equations to be satisfied by the coefficients of the spline function are

  (21)

We observe that

Table 1. Approximate solutions and absolute errors for Example 1 with h = 0.25.

also since

and the equations of (21) for reduces to

It follows that we have to determine the 5 unknowns in Equation (3), subject to the five conditions

 (22)

From (3)

(23)

(24)

Substituting (3), (23), (24) in (22)

We get the system of equations

(25)

Solving these we get

also we have

Substituting all these values in Equation (3) we get the spline approximation for which is given by

(26)

where

The analytical solution of (19) with the conditions (20) is given by

           (27)

The exact value of it follows that the absolute error in the numerical approximation is found to be which is very small.

2) Solution with

The interval is divided in to 4 equal subintervals we denote the knots by where

We assume the spline function which approximates in the form is given by (12)

From (12) we have

(28)

The conditions to be satisfied by are

(29)

for from (29) we find that

Table 2. Approximate solutions and absolute errors for Example 2 with h = 0.5.

using the remaining conditions of (29) in the order,

that is taking the Equations (12), (16), (28) in (29) & by substituting the values of

We get the following system of equations

(30)

Solving (29) we get

      (31)

Also we have

Substituting these values in (12) we get the approximation.

The values of and the corresponding absolute errors at are mentioned in Table 2.

4. Conclusion

Numerical values obtained by the spline approximation have high accuracy. It has been noticed that the numerical solutions obtained are remarkably accurate and have negligible percentage errors even for values of h as large as 0.5, 1.0.

REFERENCES

NOTES

^*Corresponding author.