The Poisson equation $\nabla^{2} u=f$ in the unit square $\Omega=[0,1] \times[0,1]$, equipped with appropriate boundary conditions on $\partial \Omega$, is discretized with the nine-point formula:

$$\begin{aligned} \Gamma_{9}\left[u_{m, n}\right] &:=-\frac{10}{3} u_{m, n}+\frac{2}{3}\left(u_{m+1, n}+u_{m-1, n}+u_{m, n+1}+u_{m, n-1}\right) \\ &+\frac{1}{6}\left(u_{m+1, n+1}+u_{m+1, n-1}+u_{m-1, n+1}+u_{m-1, n-1}\right)=h^{2} f_{m, n} \end{aligned}$$

where $1 \leqslant m, n \leqslant M, u_{m, n} \approx u(m h, n h)$, and $(m h, n h)$ are grid points.

(i) Find the local error of approximation.

(ii) Prove that the error is smaller if $f$ happens to satisfy the Laplace equation $\nabla^{2} f=0 .$

(iii) Hence show that the modified nine-point scheme

$$\begin{aligned} \Gamma_{9}\left[u_{m, n}\right] &=h^{2} f_{m, n}+\frac{1}{12} h^{2} \Gamma_{5}\left[f_{m, n}\right] \\ &:=h^{2} f_{m, n}+\frac{1}{12} h^{2}\left(f_{m+1, n}+f_{m-1, n}+f_{m, n+1}+f_{m, n-1}-4 f_{m, n}\right) \end{aligned}$$

has the same smaller error as in (ii).

[Hint. The nine-point discretization of $\nabla^{2} u$ can be written as

$$\Gamma_{9}[u]=\left(\Gamma_{5}+\frac{1}{6} \Delta_{x}^{2} \Delta_{y}^{2}\right) u=\left(\Delta_{x}^{2}+\Delta_{y}^{2}+\frac{1}{6} \Delta_{x}^{2} \Delta_{y}^{2}\right) u$$

where $\Gamma_{5}[u]=\left(\Delta_{x}^{2}+\Delta_{y}^{2}\right) u$ is the five-point discretization and

$$\begin{aligned} &\Delta_{x}^{2} u(x, y):=u(x-h, y)-2 u(x, y)+u(x+h, y) \\ &\Delta_{y}^{2} u(x, y):=u(x, y-h)-2 u(x, y)+u(x, y+h) \end{aligned}$$