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Does the following lax scheme:

$$\frac{u_{j}^{n+1}-\frac{u_{j+1}^{n}+u_{j-1}^{n}}{2}}{\Delta t}+c\frac{u_{j+1}^{n}-u_{j-1}^{n}}{2\Delta x}=0$$ of $$\begin{cases} u_t+cu_x=0 & \text{for} \ (x,t)\in\mathbb{R}\times ]0,T[ \\ u(x,0)=u_0(x) \end{cases}$$ accuracy and give the order? We have the order of accuracy in space is $O(\Delta x)^2$ since we have approximated $u_x$ by $$\frac{\partial u_{j}^{n}}{\partial x}\approx\frac{u_{j+1}^n-u_{j-1}^n}{2\Delta x}$$ but I have the problem with the following term in time: $$\frac{u_{j+1}^{n}+u_{j-1}^{n}}{2}$$

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The order can be found - as usual - by Taylor development.

$$u((n+1)\Delta t,j\Delta x) - \frac 12 (u(n\Delta t,(j-1)\Delta x) + u(n\Delta t,(j+1)\Delta x))$$

$$ = u(n\Delta t,j\Delta x) + \partial_t u(n\Delta t,j\Delta x)\Delta t + O(\Delta t^2) - \frac 12 (2 u(n\Delta t,j\Delta x) + \partial_x^2 u(n\Delta t,j\Delta x) \Delta x^2 + O(\Delta x^4))$$

$$ = \partial_t u(n\Delta t,j\Delta x)\Delta t + O(\Delta t^2) - \frac 12 ( \partial_x^2 u(n\Delta t,j\Delta x) \Delta x^2 + O(\Delta x^4)).$$

Therefore, the first term of your scheme looks like $\partial_t u + O(\Delta t) + O\left(\frac{\Delta x^2}{\Delta t}\right)$. If somehow you can guarantee that the latter term does not explode, the scheme looks ok.

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  • $\begingroup$ @TZkrevskiy. Thank you for this answer. $\endgroup$ Commented May 11, 2017 at 14:15

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