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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.

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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 $\mathcal{O}(h^2+k^2)$.

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