Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 41C / b

The first stage isBecause $A_y=I\otimes T$, this consists of $m$ independent tridiagonal systems of dimension $m$. The Thomas algorithm solves each in $O(m)$ operations, for a total of $O(m^2)$.

The second stage,is multiplication by a matrix with only $O(m^2)$ nonzero entries, so it also costs $O(m^2)$. Hence one complete step uses at mostarithmetic operations.

Eliminating the intermediate vector gives the amplification matrixThe commuting real symmetric matrices admit simultaneous diagonalization in the orthonormal basis $v^{(p,q)}$ from the question. On this vector,Since the basis is orthonormal, stability in the discrete Euclidean norm is equivalent to $|c_{pq}|\le 1$ for every $p,q$.

PutThen $-\beta \le {\lambda }_j\le -\alpha$. The smallest denominator is $1+\mu \alpha$, while the largest numerator modulus occurs at an endpoint of this interval. The condition involving $1-\mu \alpha$ is automatic, and the other isFor $m>1$, this is equivalent toSince $\beta -\alpha =4\cos (\pi h)$, the exact finite-grid stability condition isFor $m=1$ there is only one mode and the method is stable for every $\mu >0$. A simple grid-independent condition, and the limiting condition as $h\to 0$, isThis is the stability limit of one-implicit-direction diffusion splitting.