Computing end-points using the Neumann condition

[hunt_mat]

TL;DR Summary

How do I compute boundary values from a Neumann condition?

Suppose I have a boundary condition $f'(0)=a$. I know the value of $f(x)$ at $x=h/2,3h/2$. Does it make sense to write:
$$f'(x)=af(x)+bf(x+h/2)+cf(x+3h/2)$$
Using Taylor series to expand, we obtain the following:
$$f'(x)=(a+b+c)f(x)+\frac{h}{2}(b+3c)f'(x)+\frac{h^{2}}{8}(b+9c)f''(x)$$
By equating coefficients, we obtain the following set of linear equations for $a,b$ and $c$:
$$a+b+c=0,\quad b+3c=\frac{2}{h},\quad b+9c=0$$
The solution of which is:
$$a=-\frac{8}{3h},\quad b=\frac{3}{h},\quad c=-\frac{1}{3h}$$
making:
$$f'(x)=-\frac{8}{3h}f(x)+\frac{3}{h}f(x+h/2)-\frac{1}{3h}f(x+3h/2)$$
Using the above equation, it is possible to obtain:
$$f(0)=\frac{3h}{8}\left(-a+\frac{3}{h}f(h/2)-\frac{1}{3h}f(3h/2)\right)$$
Is this a reasonable approach? I've tested it with a few functions and it appears to be accurate to $10^{-4}\%$

 

[pasmith]

You are using $a$ as both the known value of $f'(0)$ and as the coefficient of $f(0)$ in your approximation to $f'(0)$. Please don't use the same symbol to mean two different things!

Generally one would treat $f(0)$ as one of the unknowns, but instead of enforcing the DE at this point one instead requires that the appropriate one-sided approximation to the derivative should equal the prescribed value of the derivative on the boundary. In an explicit time-marching scheme this reduces to your given expression.

 

[hunt_mat]

This is part of finite volume scheme, and I'm reducing the set of PDEs into a set of ODEs, a method of lines approach, and then I'm using ODE15s. So I'm converting averages within cells to values on the cell boundary.