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1.3 Forcing Functiontex2html_wrap_inline6481

Now, let us consider the example in one-dimensional case with the external forcing function is a jumped discontinuous function.

eqnarray2272

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

eqnarray2274

and

eqnarray2276

Again we use the same Green's function (1.1.12) to find the exact solution for the problem (1.3.1). However, we have a jumped discontinuous forcing function in this example, we need to divide it into two interval of x when finding the exact solution, that is
if tex2html_wrap_inline6493 , then

displaymath6495

On the other hand, if tex2html_wrap_inline6497 , then

displaymath6499

Therefore, the exact solution for the problem (1.3.1) is

eqnarray2288

Again, using the same notations as (1.1.13) and (1.1.14), we can obtain the same approximation for the differential operator as (1.1.15), but the forcing function is jumped discontinuous at the nodal point x = 1/2. However, we need to assign a suitable function value for the nodal point x = 1/2 in order to construct the difference equation for this problem. In this report, we have tried two different possibilities. The numerical solutions from these two possibilities will be compared with the convergence and the rate of convergence of the numerical solutions. In the first possibility, we assign the function value at the nodal point x = 1/2 to be 1. Therefore, the difference equation of the problem (1.3.1) should be

eqnarray2292

So that we can form a matrix equation to find the nodes tex2html_wrap_inline6411 , that is

eqnarray2294

where A is an order (n-1) square matrix, u and b are tex2html_wrap_inline6421 column matrices, such that

eqnarray2296

TABLE 3.1

 
h=1/4
.57192374x10-1
.42789110x10-3
.70207891x10-6
.29525648x10-9
h=1/8
.28810133x10-1
.54100706x10-4
.22225996x10-7
.23383332x10-11
h=1/16
.14432117x10-1
.67820360x10-5
.69682763x10-9
.17643433x10-13
h=1/32
.72194494x10-2
.84836423x10-6
.21792955x10-10
-.18219913x10-14
h=1/64
.36101489x10-2
.10606460x10-6
.67926480x10-12
 
h=1/128
.18051275x10-2
.13258669x10-7
   
h=1/256
.90257036x10-3
     

TABLE 3.2

r(1)
.50374081
.12643569
.031657404
.0079196677
r(2)
.50093892
.12535947
.031351919
.0075453031
r(3)
.50023496
.12508990
.031274528
 
r(4)
.50005876
.12502248
.031169008
 
r(5)
.50001469
.12500560
   
r(6)
.50000367
     

Table 3.1 and Table 3.2 show the numerical solutions error and the ratio of two consecutive errors at nodal point x = 1/2 respectively. We see that the numerical solutions are convergent but the numerical solution error is relatively large, though the step size is h = 1/256, the error is still tex2html_wrap_inline6531 . Obviously, the rate of convergence is slower than the problems (1.1.1) and (1.2.1). From Table 3.2, we obtain the asymptotic error expansion at this point is

eqnarray2298

The extrapolating technique at this nodal point is still effective, we get a numerical solution with error tex2html_wrap_inline6533 through extrapolating from the three solutions corresponding to h = 1/4, 1/8 and 1/16. This error is less than the numerical solution error tex2html_wrap_inline6531 corresponding to h = 1/256. We also investigate the numerical solution error at other nodal point, say x = 0.25, we see that the rate of convergence of the numerical solution is also slow or even slower than the solution at x = 0.5 [See Appendix C.3a-C.3d]. From the numerical results, we may suggest the asymptotic error expansion to be

eqnarray2300

At the nodal point x = 3/4, we have a similar result as at the nodal point x = 1/4. If it assumes that we also have the same asymptotic error expansion as (1.3.9), the extrapolated solutions are also convergent and agree with the expansion even though the second column of data do not. On the other hand, if we assigned the function value at the nodal point x = 1/2 to be -1, we will obtain the same expansion as the above situation at x = 1/2,. But the characteristic of the nodal point x = 1/4 and x = 3/4 are vice versa.

In the second situation, we assign the function value at the nodal point x = 1/2 to be 0. The change in the matrix equation (1.3.6) is only the (n/2)th entry of the column vector b which should be changed to 0. That is

eqnarray2302

At the nodal point x = 1/4, we have the following numerical results.

TABLE 3.3

 
h=1/4
-.15334056x10-3
-.31306321x10-6
-.19408943x10-9
-.33369141x10-13
h=1/8
-.38569937x10-4
-.19748410x10-7
-.30654958x10-11
-.38163916x10-16
h=1/16
-.96572955x10-5
-.12371495x10-8
-.47937349x10-13
-.13877788x10-15
h=1/32
-.24152517x10-5
-.77366783x10-10
-.88470897x10-15
-.95062846x10-15
h=1/64
-.60387096x10-6
-.48362529x10-11
-.95062846x10-15
 
h=1/128
-.15097137x10-6
-.30315681x10-12
   
h=1/256
-.37743069x10-7
     

TABLE 3.4

r(1)
.25153122
.063081221
.015794244
.0011436889
r(2)
.25038401
.062645526
.015637715
 
r(3)
.25009608
.062536324
.018455526
 
r(4)
.25002402
.062510715
   
r(5)
.25000601
.062684233
   
r(6)
.25000151
     

Table 3.3 shows the rate of convergence of the numerical solutions at nodal point x = 1/4 is larger than that in the first situation. From the Table 3.4, we obtain the same asymptotic error expansion as (1.1.19). And we also have a good extrapolating result, with h = 1/256, the finite difference solution gives the error ( tex2html_wrap_inline6567 ) which is larger than the error ( tex2html_wrap_inline6569 ) of the extrapolated solution obtained from the three solutions corresponding to h = 1/4, 1/8 and 1/16. And the solution error at the nodal point x = 1/2 is very small [See Appendix C.3e-C.3h]. Though we cannot find an asymptotic error expansion at this point, the result is good enough.


nextuppreviouscontents
Next:1.4 Forcing Function Up:1 One-dimensional ProblemsPrevious:1.2 Forcing Function
Cheung Sau Hung

Wed Sep 15 10:03:39 HKT 1999