Re: Working on bvp4c

[c.]

On 23 Aug 2016, at 11:29, lakerluke <address@hidden> wrote:

> Maybe I've misunderstood this but from page 6, f_{i - 1/2} is given by odefun
> evaluated at the point (*): 
> 
> [ (x_{i - 1} + h/2) , 0.5*(y_{i - 1} + y_{i}) - (h / 8) * (f_{ i } - f_{ i -
> 1}) ]
> 
> However,in order to evalue f at this point requires us to know how to
> evaluate the components of f at x = (x_{i - 1} + h/2), which we do not know.
> 
> Take for example demo 1 given here:
> 
> http://octave.sourceforge.net/odepkg/function/bvp4c.html
> 
> f = @(t,u) [ u(2); -abs(u(1)) ];
> 
> Seeing as u is a function of t, in order to evaluate f at (*) do we not need
> to know u as a function of t (e.g. the solution) in order to evaluate it?
> Or, seeing as in this example, f does not explicitly depend on t, do we just
> evaluate the u components of f at the second component of the point (*) -
> effectively treating f = f(u)?

If I understand correctly what formula you refer to, it is the one on line 5 at 
page 306 of the article.

In that formula, f is a function of two inputs, both of which are computable at 
the current step in the algorithm so I really don't understand what your 
question is about.

Maybe it helps if you consider the formula programmatically, i.e. divide the 
evaluation of f_{i - 1/2} into 5 steps:

1) f_{i}       = f (x_{i}, y_{i}, p) 

2) f_{i - 1}   = f (x_{i - 1}, y_{i - 1}, p) 

3) arg1        = (x_{i - 1} + h_{i - 1} / 2)

4) arg2        = 0.5 * (y_{i - 1} + y_{i}) - ...
                 (h_{i - 1} / 8) * (f_{i} - f_{i - 1});

5) f_{i - 1/2} = f (arg1, arg2, p)

at each of the 5 steps all of the quantities appearing on the RHS are known so 
the quantitities on the LHS are computable.
Which of the above steps is unclear to you?

c.