Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 40C (Numerical Analysis)

The Gershgorin circle theorem states that every eigenvalue $\lambda$ of a complex matrix $A=(a_{ij})$ lies in at least one diskTo prove it, choose a nonzero eigenvector $x$ for $\lambda$ and an index $i$ such that $|x_i|={\max }_j|x_j|>0$. The $i$th component of $Ax=\lambda x$ givesTaking absolute values and using $|x_j|\le |x_i|$ yieldsDivision by $|x_i|$ proves the theorem.

Sample the diffusion coefficient at edge midpoints and writeCentered differences of the two fluxes give the conservative finite-difference stencilFor example, the $x$ part isTaylor expansion about $(ih,jh)$ shows that the odd powers cancel between the two face fluxes and that the result equals ${\partial }_x(a{u}_x)+O(h^2)$. The same calculation in $y$ gives total truncation error

Let $L_h$ be the spatial matrix, so the Forward Euler method isThe conservative use of shared edge coefficients makes $L_h$ a real symmetric matrix. In an interior row, put $s=\alpha +\beta +\gamma +\delta <4a_{\max }$. Its diagonal entry is $-s/h^2$, and the sum of the absolute off-diagonal entries is $s/h^2$. Boundary rows have no larger disk because homogeneous Dirichlet data remove some unknown neighbors.

The Gershgorin circle theorem and symmetry therefore place every eigenvalue of $L_h$ inIf $\mu =k/h^2$, each amplification eigenvalue is $1+k\lambda$. It lies in $[-1,1]$ wheneverorSince the amplification matrix is symmetric, bounding all its eigenvalues in modulus by one bounds its discrete Euclidean operator norm by one. This proves the stated stability condition and is the Gershgorin stability bound for a variable-coefficient diffusion stencil.