[Math] Accuracy of Lax-Wendroff scheme

Question

Consider the Lax-Wendroff scheme
$$\frac{u_j^{n+1}-u_j^n}{\Delta t}+\frac{u_{j+1}^n-u_{j-1}^n}{2\Delta x}+\frac{\Delta t}{2}\frac{2u_j^n-u_{j+1}^n-u_{j-1}^n}{\Delta x^2}=0$$
for differential equation
$$\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}=0.$$ It's known that this scheme is second order accurate in time, $\mathcal{O}(\Delta t^2)$. How is this true since the scheme uses forward difference for time derivative? Recall that
$$\frac{\partial u}{\partial t} \Big|_j^n=\frac{u_j^{n+1}-u_j^n}{\Delta t}+\mathcal{O}(\Delta t).$$ Using this observation, shouldn't this scheme be first order accurate in time, $\mathcal{O}(\Delta t)$? I have read several books on numerical analysis and the authors vaguely explain this. They use the words such as "clearly" or "obviously" instead of elaborate the derivation of the accuracy.

Answer 1

You can derive the consistency order of the scheme by using the Taylor expansion around (in this case) $u_j^n$. Use the expansion up to the second derivative in time (including it) and up to the third derivative in space and then note that, from the given equation, $$ u_t=-u_x ~ \text{and} ~ u_{tt}=u_{xx}, $$ which will help to get rid of some of the terms in the expansions. The resulting approximation order should be $\mathcal{O}(h^2+k^2)$, where $h$ is spatial step and $k$ is transient step.

If you follow the derivation of the method, you will see that it uses the Taylor expansion of order $k^3$ ($k^2$ approximation power) and $h^2$ schemes for replacing the first and second derivatives in time by spatial derivatives, which clearly indicates that the method must be $\mathcal{O}(h^2+k^2)$.