Graduate Computing end-points using the Neumann condition

[hunt_mat]

TL;DR

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:
<br /> f&#039;(x)=af(x)+bf(x+h/2)+cf(x+3h/2)<br />
Using Taylor series to expand, we obtain the following:
<br /> f&#039;(x)=(a+b+c)f(x)+\frac{h}{2}(b+3c)f&#039;(x)+\frac{h^{2}}{8}(b+9c)f&#039;&#039;(x)<br />
By equating coefficients, we obtain the following set of linear equations for a,b and c:
<br /> a+b+c=0,\quad b+3c=\frac{2}{h},\quad b+9c=0<br />
The solution of which is:
<br /> a=-\frac{8}{3h},\quad b=\frac{3}{h},\quad c=-\frac{1}{3h}<br />
making:
<br /> f&#039;(x)=-\frac{8}{3h}f(x)+\frac{3}{h}f(x+h/2)-\frac{1}{3h}f(x+3h/2) <br />
Using the above equation, it is possible to obtain:
<br /> f(0)=\frac{3h}{8}\left(-a+\frac{3}{h}f(h/2)-\frac{1}{3h}f(3h/2)\right)<br />
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&#039;(0) and as the coefficient of f(0) in your approximation to f&#039;(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.