Example 3.  Solve   .  
3 (a).   Use the boundary values    and  .   

Solution 3 (a).

Form the linear operator, the differential equation and the boundary values.

The first function will be chosen to be the straight line between the boundary points.

The other n function will be chosen to be zero at the endpoints.

The residual is formed by substituting    into  .

Galerkin's requirement is that the inner product of the residual with is zero.

Form the equations      for  .  

We must solve for the coefficients   .

Use    to form the Galerkin solution.

We are done.  

Aside.  We can use Mathematica to find the analytic solution.  This is just for fun !

Plot the analytic solution.

Plot both the analytic and Galerkin solution.

Check out the difference between the analytic solution and the Galerkin solution.

So the Galerkin solution is very accurate.

Warning.  We must proceed with caution when using Galerkin's method because the linear system might be ill conditioned.  

    The condition number of the above system can be determined by Mathematica.

Fact.  Given the linear system  .  If     are input with machine precision then a bound for the error in the computed solution    is given by

        

where is machine epsilon for the computer.  The computed solution    loses about    decimal digits of accuracy relative to precision of input.

Caveat.  Mathematica uses extended precision sixteen digit numbers, the solution    retains most of this accuracy.  

We are really done.  

Aside.  We can calculate the coefficients    by directly setting up the matrix   vector .  This is just for fun !  

The boundary values    are used to form    and there are equations to solve  

        for  .  

Since the linear operator is    we have    which must be entered into the integrals on the right hand side.

        for  .  

The matrix form of the solution is  

This is the same as we obtained above.  

(c) John H. Mathews 2005