Page 319 - Applied Numerical Methods Using MATLAB

308 ORDINARY DIFFERENTIAL EQUATIONS
Needless to say, you should increase the dimension of the matrix A
to N and move the x N term on the right-hand side of the (N − 1)th row
back to the left-hand side by incorporating the corresponding statement
into the for loop. What you have to do with “bvp2m_fdf()” for this
job is as follows:
ž Make the declaration part have the boundary coefficient vector cf =
[c1 c2 c3] instead of the final position (xf) as the seventh input
argument.

function [t,x] = bvp2m_fdf(a1,a0,u,t0,tf,x0,cf,N)
ž Replace some statement by A = zeros(N,N).
ž Increase the last index of the for loop to N-1.
ž Replace the statements corresponding to the (N − 1)th row
equation by

A(N,N-1:N) = [-cf(2) cf(1)*h + cf(2)]; b(N) = cf(3)*h;
which implements Eq. (P6.7.4).
ž Modify the last statement arranging the solution as

x = [x0 trid(A,b)’]’;

Now that you have the routine “bvp2m_fdf()” of your own making,
don’t hesitate to try it on the following problem:

2 2

x (t) + x (t) − x(t) = 0 with x(1) = 5,x(2) + x (2) = 3
t t 2
(P6.7.5)
For this job, you only have to modify one statement of the program
“do_fdf.m” (Section 6.6.2) into
[t,x] = bvp2m_fdf(a1,a0,u,t0,tf,x0,[1 1 3],N);

You might need to increase the number of segments N to improve the
accuracy of the numerical solution. If you run it to obtain the same
solution as depicted in Fig. 6.9, be happy with it.
6.8 BVP with Mixed-Boundary Conditions II

Suppose the boundary condition for a second-order BVP is given as

c 01 x(t 0 ) + c 02 x (t 0 ) = c 03 (P6.8.1a)

c f 1 x(t f ) + c f 2 x (t f ) = c f 3 (P6.8.1b)
Consider how to modify the MATLAB routines “bvp2m_shoot()”and
“bvp2m_fdf()” so that they can accommodate this kind of problems.