# finite difference methods and global error

I was going through my notes on different finite difference methods and came across something I don't quite understand. I have code that will calculate an approximate solution we can call this $U_{nm}$ that I define on a grid using $h$ and $dt$ for the change in $x$ and time, respectively. Now I have written down the global error is just: $$e_{nm} =|U_{nm} - u(x_n,t_n)|$$ where $u(x_n,t_n)$ is the exact solution evaluated at the gird points. From there we can calculate our rate of convergence.

So I, perhaps naively, just assumed I could take the solution I calculate and subtract the exact solution at every point take the absolute value. However, I have written something about actually just using $e_{nm}$ to calculate the initial values as well as boundary conditions and then plugging them back into the finite difference method to calculate all the grid points for all of $e_{nm}$.

Is this correct or did I perhaps not fully understand what my instructor was saying?

(Note we are working with forward, backward and crank-nicolson methods)

• Would it be best for me to move it and just delete this question from here? Feb 26 '12 at 18:21
• I flagged your question hoping it gets moderators attention & they would migrate the question to scicomp.SE.
– user182
Feb 26 '12 at 18:24
• @user25783: I take your comment as indicating that you want it moved. So here it goes.
– Willie Wong
Feb 26 '12 at 23:27
• this would work if you have the exact solution available... otherwise, you would have to find an upper bound on the error $e_{mn}$ in terms of the truncation error (from the finite difference scheme that you chose, i.e. crank-nicolson).
– Paul
Mar 16 '12 at 13:28

However, I have written something about actually just using $e_{nm}$ to calculate the initial values as well as boundary conditions and then plugging them back into the finite difference method to calculate all the grid points for all of $e_{nm}$.