Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 40A (Numerical Analysis)

Take $b$ to be the vector of $f_{ij}$ and absorb $h^{-2}$ into $A$. Its diagonal entry is $-10/(3h^2)$, each axial-neighbour entry is $2/(3h^2)$ and each diagonal-neighbour entry is $1/(6h^2)$. Since these weights agree in both directions, $A$ is a symmetric matrix.

Extend a grid vector $u$ by zero to the whole integer lattice. Let ${\mathcal{E}}_a$ contain unordered axial-neighbour pairs and ${\mathcal{E}}_d$ unordered diagonal-neighbour pairs. Only finitely many differences involving nonzero values contribute. Each lattice vertex has total neighbour weight $4(2/3)+4(1/6)=10/3$, so expanding the squared differences givesIt is nonpositive. If it vanishes, every axial difference vanishes. The axial lattice is connected and $u$ is zero outside the finite interior square, forcing $u=0$. Thus $u^{T}Au<0$ for every nonzero $u$: $\boxed{A\text{ is negative definite}}$.

Reordering the grid replaces $A$ by $P^{T}AP$ for a permutation matrix $P$. This preserves symmetry and the sign of the quadratic form, proving the assertion for any ordering. Multiplying the whole system by $h^2$ merely rescales $A$ by a positive number and has the same conclusions.

Let $D=-10I/(3h^2)$ be the diagonal part of $A$. The weighted Jacobi method isEquivalently it is $u^{(r+1)}=u^{(r)}+\omega D^{-1}(b-A{u}^{(r)})$. In grid form, all neighbours on the right are taken from the old iterate:The boundary values stay zero at every step. The iteration matrix is $T_{\omega }=I-\omega D^{-1}A$.

For $\omega =1$, the Jacobi method iteration matrix $T=I-D^{-1}A$ has zero diagonal, nonnegative axial entries $1/5$, and diagonal-neighbour entries $1/20$. Its row sums are at most one; they are strictly below one at grid points next to the Dirichlet boundary. Its graph is connected by the positive axial entries.

To prove the strict spectral radius bound, let $Tz=\lambda z$ with $z\ne 0$ and choose a component of maximum magnitude $M>0$. Thenso $|\lambda |\le 1$. If equality held, every positive-weight neighbour of a maximal component would also have modulus $M$, and that row would have sum one. Propagating along axial edges reaches a boundary-adjacent row, whose sum is strictly less than one, a contradiction. This also covers a one-point interior grid, for which $T=0$. HenceThe standard convergence theorem for a stationary linear iteration says it converges from every initial vector exactly when its iteration matrix has spectral radius below one. Since $A$ is invertible by (i), the error satisfies $e^{(r)}=T^r{e}^{(0)}\to 0$ andWeak row diagonal dominance alone would not prove convergence; the connection to a strictly deficient boundary row supplies the required strictness.