1.1 Forcing Function 

  Now, let us consider the following example

According to (3.2) and (3.3), we decompose our problem into the following problems,

and

The first problem above has the general solution

where 
It subjects to the prescribed boundary conditions, leads to  . Hence,  . Next, we need to find  by the method of Green's function. We first observe from (3.17a) that is either  or  . Then, from the definition of delta function, we could have

If  and  are linearly independent solutions of the above homogeneous differential equation, each of them satisfies one homogeneous condition and is selected in the way that

Then, we could have

Hence, the Green's function has the form

where u and v are functions to be determined. Imposing conditions (3.17c and 3.17d), it follows that the unknown functions u and v must be chosen such that

The simultaneous solution yields

where  is the Wronskian function. Since  , the Green's function is , accordingly,

Now, the external disturbance is a continuous sine function. Using the integration by parts, therefore, the solution of  is

So, the exact solution of the problem (1.1.1) is

We are going to use the method of finite difference to investigate the problem (1.1.1). Under the consideration of the problem (1.1.1), we divide the interval [0,1] into n equal parts, then the step size is h = 1/n and we have (n+1) node points. That is, for each node point,

And let the corresponding function value be

Therefore, the problem (1.1.1) can be approximated by

Then we could have

So that we can form a matrix equation to find the other nodes , that is,

where A is an order (n-1) square matrix, u and b are  column matrices, such that

TABLE 1.1

h=1/4	.26821964x10-4	-.16287549x10-6	-.10167418x10-9	-.17034985x10-13
h=1/8	.65833344x10-5	-.10275038x10-7	-.16054241x10-11	.28449465x10-15
h=1/16	.16381273x10-5	-.64369495x10-9	-.2480654610-13	-.19428903x10-15
h=1/32	.40904906x10-6	-.40254189x10-10	-.57592819x10-15	.23314684x10-14
h=1/64	.10223207x10-6	-.25164315x10-11	.22828961x10-14
h=1/128	.25556131x10-7	-.15513285x10-12
h=1/256	.63889164x10-8

 
 

TABLE 1.2

r(1)	.24544565	.063085230	.015789890	-.016700611
r(2)	.24882943	.062646480	.015451709
r(3)	.24970529	.062536126	.023216783
r(4)	.24992619	.062513532
r(5)	.24998154	.061647953
r(6)	.24999545

We can compare the solution error at different nodal point. Table 1.1 shows the solution error at the point x = 1/2, where we denote the numerical solution,  , after k steps of extrapolation. Table 1.2 shows the ratio between two consecutive errors according to (5.2). The numerical results completely agree with the theoretical result in paper [1]. From the first column of the data of Table 1.2, we can see that the ratio tends to 0.25 gradually, that is,  can be determined from (5.1), we take . Using (5.3), we can compute the first step of extrapolation and the corresponding errors as shown in the second column of the data of the Table 1.1. Repeat the same procedure, we can take  . Therefore, we could have the asymptotic error expansion is

We also see that extrapolating from the three solutions corresponding to h=1/4, 1/8 and 1/16, we get a numerical solution with error  , which is smaller than the numerical solution error  corresponding to h=1/256. At other nodal points, we find that we have a similar results as Table 1.1 and Table 1.2, we also find that we have an identical asymptotic error expansion as (1.1.19) and have the similar improvement on the numerical solution accuracy by using extrapolation [See Appendix A.1a-A.1c].

Wed Sep 15 10:03:39 HKT 1999