For the advection equation $v_t + v_x = 0$, For practice, i want to
show the scheme FTCS$$ u_i^{n+1} = u_i^n – \frac{\Delta t }{2 \Delta x } (u_{i+1}^n –
u_{i-1}^n ) $$is consistent and moreover I want to find the truncation error.
Let $U_i^n = v(i\Delta x, n \Delta t)$ be the ${\bf exact}$ solution to the pde. By definition I need to show that $U_{i}^n – u_i^n $ tends to zero as $\Delta t, \Delta x \to 0$. Im kind of unsure how to use this definition, but does it mean that I need to check the following:
$$ \frac{ f_i^{n+1} – f_i^n }{\Delta t } + \frac{ f_{i+1}^n – f_{i-1}^n }{2 \Delta x } =0$$
where $f_i^n = U_i^n – u_i^n$. This reduces to
$$ \frac{ U_i^{n+1} – U_i^n }{\Delta t } + \frac{ U_{i+1}^n – U_{i-1}^n }{2 \Delta x } =0$$
since $u_i^n$ satisfies the scheme and thus gives zero. Now we can use taylor
$$ U_i^{n+1} = U_i^n + (U_t)_i^n \Delta t + (U_{tt})_i^n \Delta^2 t /2 + O(\Delta^3 t ) $$
Similarly,
$$ U_{i+1}^{n} = U_i^n + (U_x)_i^n \Delta x + (U_{xx})_i^n \Delta^2 x /2 + O(\Delta^3 x ) $$
$$ U_{i-1}^{n} = U_i^n – (U_x)_i^n \Delta x + (U_{xx})_i^n \Delta^2 x /2 – O(\Delta^3 x ) $$
After cancellation the above reduces to
$$ (U_t)_i^n + (U_{tt})_i^n \frac{\Delta t }{2} + O(\Delta t^2) + (U_x)_i^n + O(\Delta^2 x) = 0$$
since $U$ satisfies pde we have
$$ (U_{tt})_i^n \frac{\Delta t }{2} + O(\Delta t^2) + O(\Delta^2 x) = 0 $$
but here is where I get stuck. How do I prove that consistency? Can someone give me some criticism to my solution? Am I on the right track?
Best Answer
Following the steps in this post, we obtain the following expansions from second-order Taylor series of space and time: \begin{aligned} U_{i}^{n+1} &= u - p\Delta x\,u_x + \tfrac{1}{2}(p\Delta x)^2 u_{xx} + O(\Delta x^3) \\ U_{i}^{n} &= u \\ U_{i\pm 1}^{n} &= u \pm \Delta x\,u_x + \tfrac{1}{2} \Delta x^2 u_{xx} + O(\Delta x^3) \end{aligned} where $p = \Delta t/\Delta x$ is the Courant number, and $U_{i+I}^{n+N}$ is defined as $(x,t) \mapsto u(x+I\Delta x, t+N\Delta t)$. Here, $u$ is a smooth solution of the advection equation $u_t + u_x = 0$. The finite difference operators become \begin{aligned} \frac{U_i^{n+1} - U_i^{n}}{\Delta t} &= -u_x + \tfrac{1}{2}pu_{xx}\Delta x + O(\Delta x^2) \\ \frac{U_{i+1}^{n} - U_{i-1}^{n}}{2\Delta x} &= u_x + O(\Delta x^2) \end{aligned} By summing up these two terms and taking the limit $\Delta x \to 0$, we have shown that the method is consistent. The local truncation error is $\tfrac{1}{2}pu_{xx}\Delta x + O(\Delta x^2)$. Hence, this method is first-order accurate in space and time (at a fixed Courant number). But as mentioned in the comments, it is useless due to severe stability issues.