Taylor series and finite differences

To formally analyse the error associated with a finite difference approximation for a derivative, one expands each possible term of the approximation in a Taylor series (TS) so that Dapprox = Taylor series. You then would take the difference between the approximation (LHS) and the derivative term from the TS (on RHS), and you are left with the 'error'. As h gets small, the biggest term of this series will dominate the error. You now have an explicit coefficient for E ~ O(hⁿ).

Replacing derivatives in differential equations with finite difference approximations replaces the original diff-eq problem with a system of algebraic equations. To analyse the error in this case, we need the concepts of "local truncation error" at each grid point, as well as some handle on the "global error" over the entire mesh.

Local truncation error (LTE)

LTE is a measure of how well a solution to a particular ODE satisfies the finite difference approximation of that ODE.

The concept of LTE is illustrated with the following example. Suppose we replace the 2nd derivative of the 1D Poisson equation,

\frac{d^2 u}{dx^2} = f(x)

with the centered (2nd order, as can be verified by above discussion) finite different approximation,

\frac{1}{h}[\frac{u(x+h)-u(x)}{h} - \frac{u(x-h)-u(x)}{-h}] = \frac{1}{h^2}[u(x+h)+u(x-h)-2u(x)]

(which is easily interpreted as the finite difference version of the derivative of the derivative). Using this, and replacing the function f and the (approximate) solution u with their discrete forms, we have

\frac{1}{h^2}[u(x_i+h)+u(x_i-h)-2u(x_i)] = f(x_i)

This is a system of equations over a 1D interval. As such, it can be written as a matrix equation,

A {\vec{u}}=\vec{x}

where {\vec{u}} is the solution vector from this finite-difference scheme.

Now if we replace the approximate solution with the exact solution in the discrete equation above, the left and right hand sides in general will not be exactly the same. This difference is the LTE, and can be written as,

\tau = \frac{1}{h^2}[u(x_i+h)+u(x_i-h)-2u(x_i)] - f(x_i)