 Solution

With zero boundary values eliminated, the diagonal of $L_h$ is $-4I/h^2$. The Jacobi method is thereforewhich averages the four neighboring components. The sine addition identity shows that the supplied product vectors are eigenvectors, withThe discrete sine vectors form a basis, so these are all eigenvalues. Their largest modulus isattained at the lowest and highest sine frequencies. Thus the asymptotic error reduction factor per iteration is $\cos (\pi h)$. If the convergence rate is expressed logarithmically, $R=-\log \rho (J)\sim {\pi }^2{h}^2/2$ as $h\to 0$. A fixed error reduction therefore takes $O(h^{-2})$ Jacobi sweeps. For the special grid $m=2$, $J=0$ and one sweep solves its single interior equation.