We consider the boundary problem \begin{align*}-&y''(x)=0 \ \ \text{ for } \ x\in \Omega:=(0,\pi) \\ &y(0)=y(\pi)=0\end{align*} Let's pick the grid size $n=5$.
I want to apply the 2-grid method.
We get the linear system \begin{equation*}A_hu_h=f_h, \ \ \ A_h:=\frac{1}{h^2}\begin{bmatrix}2 & -1 & 0 & 0 \\ -1 & 2 & -1& 0 \\ 0 & -1& 2 & -1 \\ 0 & 0& -1 & 2\end{bmatrix}=\frac{25}{\pi^2}\begin{bmatrix}2 & -1 & 0 & 0 \\ -1 & 2 & -1& 0 \\ 0 & -1& 2 & -1 \\ 0 & 0& -1 & 2\end{bmatrix}, \ \ f_h:=\begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}\end{equation*}
We apply twice the damped Jacobi method with the relaxation factor $\omega=\frac{2}{3}$.
We have the matrix $J = I-\frac{h^2}2 A_h=\begin{bmatrix}1 & 0 & 0 &0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}-\begin{bmatrix}1 & -\frac{1}{2} & 0 & 0 \\ -\frac{1}{2} & 1 & -\frac{1}{2}& 0 \\ 0 & -\frac{1}{2}& 1 & -\frac{1}{2} \\ 0 & 0& -\frac{1}{2} & 1\end{bmatrix}=\begin{bmatrix}0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2}& 0 \\ 0 & \frac{1}{2}& 0 & \frac{1}{2} \\ 0 & 0& \frac{1}{2} & 0\end{bmatrix}$.
For $\omega=\frac{2}{3}$ we get the following damped version: $J(\omega) = (1-\omega)I + \omega J= \frac{1}{3}I + \frac{2}{3}J=\begin{bmatrix}\frac{1}{3} & 0 & 0 &0 \\ 0 & \frac{1}{3} & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & \frac{1}{3}\end{bmatrix}+\begin{bmatrix}0 & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & 0 & \frac{1}{3}& 0 \\ 0 & \frac{1}{3}& 0 & \frac{1}{3} \\ 0 & 0& \frac{1}{3} & 0\end{bmatrix}=\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}& 0 \\ 0 & \frac{1}{3}& \frac{1}{3} & \frac{1}{3} \\ 0 & 0& \frac{1}{3} & \frac{1}{3}\end{bmatrix}$
The damped Jacobi method is then: $$v^{(i+1)} = J(\omega)v^{(i)} + \frac\omega 2{h^2}f=\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}& 0 \\ 0 & \frac{1}{3}& \frac{1}{3} & \frac{1}{3} \\ 0 & 0& \frac{1}{3} & \frac{1}{3}\end{bmatrix}v^{(i)}$$
Let's consider the starting vector $v^{(0)}=\begin{bmatrix}1\\ 1\\ 1\\ 1\end{bmatrix}$.
Then we get the following: $$v^{(1)} = J(\omega)v^{(0)} + \frac\omega 2{h^2}f=\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}& 0 \\ 0 & \frac{1}{3}& \frac{1}{3} & \frac{1}{3} \\ 0 & 0& \frac{1}{3} & \frac{1}{3}\end{bmatrix}v^{(0)}=\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}& 0 \\ 0 & \frac{1}{3}& \frac{1}{3} & \frac{1}{3} \\ 0 & 0& \frac{1}{3} & \frac{1}{3}\end{bmatrix}\begin{bmatrix}1\\ 1\\ 1\\ 1\end{bmatrix}=\begin{bmatrix}\frac{2}{3}\\ 1\\ 1\\ \frac{2}{3}\end{bmatrix}$$ $$v^{(2)} = J(\omega)v^{(1)} + \frac\omega 2{h^2}f=\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}& 0 \\ 0 & \frac{1}{3}& \frac{1}{3} & \frac{1}{3} \\ 0 & 0& \frac{1}{3} & \frac{1}{3}\end{bmatrix}v^{(1)}=\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}& 0 \\ 0 & \frac{1}{3}& \frac{1}{3} & \frac{1}{3} \\ 0 & 0& \frac{1}{3} & \frac{1}{3}\end{bmatrix}\begin{bmatrix}\frac{2}{3}\\ 1\\ 1\\ \frac{2}{3}\end{bmatrix}=\begin{bmatrix}\frac{5}{9}\\ \frac{8}{9}\\ \frac{8}{9}\\ \frac{5}{9}\end{bmatrix}$$
The residual is then $$r_h:=f_h-A_hv^{(2)}=-\frac{25}{\pi^2}\begin{bmatrix}2 & -1 & 0 & 0 \\ -1 & 2 & -1& 0 \\ 0 & -1& 2 & -1 \\ 0 & 0& -1 & 2\end{bmatrix}\begin{bmatrix}\frac{5}{9}\\ \frac{8}{9}\\ \frac{8}{9}\\ \frac{5}{9}\end{bmatrix}=-\frac{25}{\pi^2}\begin{bmatrix}\frac{2}{9}\\ -\frac{1}{3}\\ \frac{1}{3}\\ \frac{2}{9}\end{bmatrix}=\begin{bmatrix}-\frac{50}{9\pi^2}\\ \frac{25}{3\pi^2}\\ -\frac{25}{3\pi^2}\\ -\frac{50}{9\pi^2}\end{bmatrix}$$
We go to the coarser grid: $$r_{2h}=I_h^{2h}r_h=\frac{1}{4}\begin{bmatrix}1 & 2 & 1 & 0 \\ 0 & 0 & 1 & 2\end{bmatrix}\begin{bmatrix}-\frac{50}{9\pi^2}\\ \frac{25}{3\pi^2}\\ -\frac{25}{3\pi^2}\\ -\frac{50}{9\pi^2}\end{bmatrix}=\begin{bmatrix} \frac{25}{36\pi^2}\\ -\frac{175}{36\pi^2}\end{bmatrix}$$
Our new problem on the coarse grid is then the following: $$A_{2h}e_{2h}=-r_{2h}$$ where \begin{align*}A_{2h}&=I_h^{2h}A_hI_{2h}^h=\frac{1}{4}\begin{bmatrix}1 & 2 & 1 & 0 \\ 0 & 0 & 1 & 2\end{bmatrix}\begin{bmatrix}\frac{50}{\pi^2} & -\frac{25}{\pi^2} & 0 & 0 \\ -\frac{25}{\pi^2} & \frac{50}{\pi^2} & -\frac{25}{\pi^2} & 0 \\ 0 & -\frac{25}{\pi^2}& \frac{50}{\pi^2} & -\frac{25}{\pi^2} \\ 0 & 0& -\frac{25}{\pi^2} & \frac{50}{\pi^2}\end{bmatrix}\frac{1}{2}\begin{bmatrix}1 & 0 \\ 2 & 0 \\ 1 & 1 \\ 0 & 2\end{bmatrix}\\ & =\frac{1}{8}\begin{bmatrix}1 & 2 & 1 & 0 \\ 0 & 0 & 1 & 2\end{bmatrix}\begin{bmatrix}\frac{50}{\pi^2} & -\frac{25}{\pi^2} & 0 & 0 \\ -\frac{25}{\pi^2} & \frac{50}{\pi^2} & -\frac{25}{\pi^2} & 0 \\ 0 & -\frac{25}{\pi^2}& \frac{50}{\pi^2} & -\frac{25}{\pi^2} \\ 0 & 0& -\frac{25}{\pi^2} & \frac{50}{\pi^2}\end{bmatrix}\begin{bmatrix}1 & 0 \\ 2 & 0 \\ 1 & 1 \\ 0 & 2\end{bmatrix}\\ & =\frac{1}{8}\begin{bmatrix} 0& \frac{50}{\pi^2}& 0& -\frac{25}{\pi^2}\\ 0& -\frac{25}{\pi^2}& 0& \frac{75}{\pi^2}\end{bmatrix}\begin{bmatrix}1 & 0 \\ 2 & 0 \\ 1 & 1 \\ 0 & 2\end{bmatrix}\\ &= \begin{bmatrix} \frac{100}{8\pi^2}& -\frac{50}{8\pi^2}\\ -\frac{50}{8\pi^2}& \frac{150}{8\pi^2}\end{bmatrix}= \begin{bmatrix} \frac{25}{2\pi^2}& -\frac{25}{4\pi^2}\\ -\frac{25}{4\pi^2}& \frac{75}{4\pi^2}\end{bmatrix}\end{align*} Now we apply again twice the damped Jacobi, right? Which starting vector do we use?
Is everythig correct so far?
Then we calculate $v^{(2)}-I_{2h}^he_{2h}$ where $e_{2h}$ is the solution of the last Jacobi method, or not?
Is this then the result, I mean the last approximation of the solution of the initial value problem?