Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 38C / b / Solution

For a linear system write $A=D+R$, where $D$ is the diagonal. The Jacobi method iteratesHere $D=(10/3)I$, and $B$ has nonnegative off-diagonal entries with row sums at most one. Rows adjacent to the boundary have sum strictly less than one; those at purely interior stencil vertices have sum one. Thus $\Vert B{\Vert }_{\infty }\le 1$.

To prove strict spectral radius below one, suppose $Bv=\zeta v$ with $|\zeta |=1$, and choose an index with $|v_i|=\Vert v{\Vert }_{\infty }>0$. Inequality forces that row sum to equal one and every linked neighbor to attain the same maximum modulus. Propagate this along the connected axial grid to a boundary-adjacent row, whose row sum is strictly less than one, a contradiction. All eigenvalues therefore have modulus strictly below one. Hence $B^k\to 0$ and the Jacobi errors tend to zero from every initial vector. The unique solution exists by positive definiteness, so $\boxed{\text{Jacobi converges for this discretization in any ordering}.}$