Example 4.  Solve   .  
4 (a).   Use the boundary values    and  .   

Solution 4 (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 appears to be good.

Caution.  We 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.

The condition number is small, we should expect a good solution.

Aside.  We can check out the matrix form for setting up the Galerkin equations.  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