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Galerkin's method

In the Galerkin's method the functions Vi(x) in (82) are defined as
equation809
Inserting (106) in (88) yields
equation811
The linear system (92) becomes

eqnarray813

Example 1: We consider the two point boundary value problem
equation825

equation828
The function g(x) and the differential operator L are determined as
equation830
The two point boundary value problem (109)-(110) has the exact solution
equation833
Next we approximate the solution of (109)-(110) with trigonometric series
eqnarray836
The boundary conditions (110) are satisfied if
equation838
Taking n=2 in (113) we get approximation u* with two terms as
equation840
Comparing (76) and (115) yields
equation842
Substituting tex2html_wrap_inline4676 and tex2html_wrap_inline4762 in (96) gives tex2html_wrap_inline4764 linear system
equation845
According to the point collocation method we choose the collocation points x1, x2 and solve (117) with respect a1 and a2. For tex2html_wrap_inline4774 and tex2html_wrap_inline4776 one obtains tex2html_wrap_inline4778and
equation858
Subdomain collocation method leads to the following solution

Taking tex2html_wrap_inline4780 (two subdomains with equal length) we can write the system (99) as
equation861
The system (119) has solutions tex2html_wrap_inline4782 and therefore
equation878
Let us use the least square method now

For the present example the system (105) reduces to
eqnarray880
Solving (121) we get tex2html_wrap_inline4784 and
equation891
Galerkin's method gives the following solution

For the considered example the system (108) reduces to
eqnarray893
Solving system (123) yields tex2html_wrap_inline4786 and therefore
equation891
Comparison of solutions:

Võrdlus

It is seen in Fig. 1., that the numerical results, obtained by four weighted residual methods are quite close to the exact solution. However, in present example only two terms in trigonometric series are considered.

Exercises

1. Solve the two point boundary value problem (109)-(110) taking n=3 and n=4. Compare results.