Consider the numerical routine:
$\frac{1}{\Delta t^2} (U_i^{n+1} -2U_i^{n} + U_i^{n-1}) – \frac{a^2}{\Delta x^2} (U_{i+1}^{n} -2U_i^{n} + U_{i-1}^{n})$
which solves the one-dimensional wave equation $u_{tt} – a^2 u_{xx} = 0$ where a $\in \mathbb{R}$ and $U_i^n$ is the numerical approximation of $u(j\Delta x,n\Delta t)$.
So my book mentions multiple times that this numerical scheme is second-order accurate. It kind of passingly mentions this with really saying how I would go about actually showing it, nor can I find any example of how to do so. The only thing it says is that "a truncation error analysis reveals second-order accuracy." I tried looking this up and again I'm having a difficult time seeing what exactly it is I need to do. I think I need to perform a taylor expansion but I don't know around which variables this needs to happen or where to evaluate it. I would appreciate it if someone could show how this is second order accurate or provide some references of where I would learn how to do so. Its something that keeps popping up.
Best Answer
Consider the spatial derivative (dropping the time index for convenience). Using the 4th order Taylor expansion we have
$$U_{i+1} = U_i + U_{x,i} \Delta x + \frac{1}{2}U_{xx,i}\Delta x^2 + \frac{1}{6}U_{xxx,i}\Delta x^3 + O(\Delta x^4), \\ U_{i-1} = U_i - U_{x,i} \Delta x + \frac{1}{2}U_{xx,i}\Delta x^2 - \frac{1}{6}U_{xxx,i}\Delta x^3 + O(\Delta x^4). $$
Adding the two equations, we get
$$U_{i+1} + U_{i-1} =2U_i + U_{xx,i}\Delta x^2 + O(\Delta x^4).$$
Rearrranging,
$$\frac{U_{i+1} -2U_i + U_{i-1}}{\Delta x^2} = U_{xx,i} + O(\Delta x^2).$$
This shows the truncation error associated with the central difference approximation of the second partial derivative is of second order. A similar result is obtained for the second partial derivative with respect to the time variable.