$$u_t + a u_x = f(x,t)$$ in $(0,1)\times (0,T)$. Apply the foward-time backward-space scheme. What is the order of accuracy in time and space?
I obtain $$U_{m}^{n+1} = U_{m}^{n} – a\frac{k}{h}\left( U_{n}^{m} – U_{m-1}^{n}\right )$$where $U_m^n$ is the approximation to $u(x_m, t_n)$, $x_m = m\cdot h$ and $t_n = n \cdot k$. The forward-time and backward-space approximations to $u_t$ and $u_x$ have $O(k)$ and $O(h)$ remainders respectively. So then in the above scheme we get $\cdots + O(k^2) + O(kh)$ after multiplying across by $k$.
I am confused with the $O(kh)$ term. Do we say the scheme is order $2$ accurate in time and order $1$ accurate in space?
Best Answer
I took a look back at my notes and found my issue. To obtain the order of accuracy / consistency, we compute the Local Truncation Error: $$R_m^n = \mathcal{L}_{h,k}u_m^n - \mathcal{F}_{h,k}$$ Here, $\mathcal{L}_{h,k}:= k^{-1}\Delta_t^{+}u_m^n + ah^{-1}\Delta_x^{-}u_m^n$ and $\mathcal{F}_{h,k}:= f_{m,n}$
Plugging the terms we eventually obtain the following expression:$$\frac{k}{2!} \partial_{tt}u_m^n + a\frac{h}{2!}\partial_{xx}u_m^n +O(k^2)+O(h^2)$$
Hence, the scheme is accurate/consistent of order $1$ in time and space.