Given a two dimensional heat equation $\displaystyle \frac{\partial u}{\partial t}=K(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2})$ solved using ADI (the alternating-direction implicit method) which is described by
$\displaystyle \frac{w_{k,j}^{n+\frac{1}{2}}-w_{k,j}^n}{\tau}=\frac{K}{2h^2}(\delta^2_xw_{k,j}^{n+\frac{1}{2}}+\delta^2_yw_{k,j}^{n})$ and
$\displaystyle \frac{w_{k,j}^{n+1}-w_{k,j}^{n+\frac{1}{2}}}{\tau}=\frac{K}{2h^2}(\delta^2_xw_{k,j}^{n+\frac{1}{2}}+\delta^2_yw_{k,j}^{n+1})$.
I am required to find the local truncation error of this method, by summing the two equations and then subtracting the second from the first I arrive at an equation which coincides with the Crank-Nicholson method with an extra term. I know that Crank-Nicholson has error $O(\tau^2+h^2)$ so must calculate the error of the extra term, which is
$\displaystyle \frac{K^2\tau}{4h^4}\delta^2_x\delta^2_y(w_{k,j}^n-w_{k,j}^{n+1})$.
I have calculated the necessary Taylor series and found
$\displaystyle \frac{1}{h^4}\delta^2_x\delta^2_yu_{k,j}^n=u_{xxyy}(x_k,y_j,t_n)+O(h^2) $ and
$\displaystyle \frac{1}{h^4}\delta^2_x\delta^2_yu_{k,j}^{n+1}=u_{xxyy}(x_k,y_j,t_{n+1})+O(h^2) $.
In the model solution the next step is to say
$\displaystyle \frac{K^2\tau}{4h^4}\delta^2_x\delta^2_y(u_{k,j}^n-u_{k,j}^{n+1})=\frac{K^2\tau^2}{4}(u_{xxyyt}(x_k,y_j,t_n)+O(\tau))+O(\tau h^2)=O(\tau^2)$.
I don't understand this final step. I can follow the first equality, but then I would have thought this was equal to $O(\tau^3)+O(\tau h^2)$. Can somebody explain why it is in fact equal to $O(\tau^2)$?
Thanks
By the way, $\delta_x^2 w_{k,j}=w_{k+1,j}-2w_{k,j}+w_{k-1,j}$.
Best Answer
The error is $O(\tau^3)+O(\tau^2)+O(\tau h^2)=O(\tau^2)+O(\tau h^2)=O(\tau^2)$ as that is of the smallest degree.