2.2 Forcing Function

Consider the second two-dimensional example, the problem is now acting a jumped discontinuous forcing function, that is

where 

Using the same technique of finding the exact solution as the previous example, the coefficient of  can be determined as follows

Therefore, the exact solution for the problem (2.2.1) is

We use the five-point finite difference method and uniform meshes similar to the above example. A similar matrix equation can be obtained, the only difference would be the column vector b, such that the entry of b would be the corresponding value of the forcing function and we set the function value at the jumped discontinuous points to be zero. The numerical solutions at the nodal point (1/4, 1/4) are as follows

TABLE 7.1

h=1/4	.95279099x10-2	.16579651x10-2	-.76531779x10-5	-.82122925x10-5
h=1/8	.55929375x10-2	.40875139x10-3	-.81424032x10-5	-.68647151x10-6
h=1/16	.30008444x10-2	.96081044x10-4	-.16184630x10-5	-.45508891x10-7
h=1/32	.15484627x10-2	.22806414x10-4	-.24212815x10-6	-.27940559x10-8
h=1/64	.78563458x10-3	.55200073x10-5	-.32710818x10-7
h=1/128	.39557729x10-3	.13554687x10-5
h=1/256	.19846638x10-3

TABLE 7.2

r(1)	.58700571	.24653799	1.0639245	.083590728
r(2)	.53654174	.23505986	.19876969	.066293925
r(3)	.51600900	.23736642	.14960376	.061395826
r(4)	.50736421	.24203750	.13509713
r(5)	.50351309	.24555560
r(6)	.50171328

From the results above, we suggest the asymptotic error expansion to be

And 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. But if we extend the idea from the Experiment (1.3), we take the function value at the jumped discontinuous points to be the average of the neighbouring nodal points in the finite difference method. The following Tables 7.3 and Table 7.4 show the numerical solutions at the same point (1/4, 1/4). We see that the results are quite better than the results from Table 7.1 and Table 7.2. And we can obtain the a symptotic error expansion

Furthermore, we obtain a similar results at any other nodal points [See Appendix G.7a-G.7f].

TABLE 7.3

h=1/4	.23771507x10-3	.65380509x10-4	.19349148x10-5	.17357554x10-7
h=1/8	.10846415x10-3	.59002644x10-5	.47319386x10-7	.10071083x10-9
h=1/16	.31541235x10-4	.41312845x10-6	.83850263x10-9	-.37014412x10-10
h=1/32	.81951552x10-5	.26606624x10-7	-.23334458x10-10	-.37767543x10-10
h=1/64	.20687438x10-5	.16410380x10-8	-.37542027x10-10
h=1/128	.51841672x10-6	.67369223x10-10
h=1/256	.12965471x10-6

TABLE 7.4

r(1)	.45627797	.090245006	.024455540	.0058021325
r(2)	.29079872	.070018633	.024455540
r(3)	.25982353	.064402789	-.027828723
r(4)	.25243497	.061677797
r(5)	.25059494	.041052812
r(6)	.25009746

Wed Sep 15 10:03:39 HKT 1999