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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 y^{x}(x)+g (x)y(x)= r(x) where g(x) and r(x) 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.

In the recent past, several authors have considered the application of cubic spline functions for the solution of two point boundary value problems. Bickley [

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

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

First, we consider the linear differential equation

With the boundary conditions

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

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

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.

Suppose the interval is divided in to “n” subintervals with knots. Starting at x_{0}, 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

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

We consider the linear fifth order differential equation

With the boundary conditions

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

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

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

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.

Consider the following fifth order linear boundary value problem

With the boundary conditions

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

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

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

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

from (6)

and

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

Solving these we get

Substituting these values in (6) we get

where h = 0.5 Therefore

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

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.

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

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

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

From (14)

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

we get the following system of equations

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

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

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.

Consider the following fifth order linear boundary value problem

Subject to

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

We observe that

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

From (3)

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

We get the system of equations

Solving these we get

also we have

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

where

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

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

The conditions to be satisfied by are

for from (29) we find that

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

Solving (29) we get

Also we have

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

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

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.