Past exam of the numerical solution of differential equations course of the University of Oxford / 2010 / A / 2 / b

Transforming the spatial shifts and putting $\xi =k\Delta x$ gives the real amplification factorFor $0\le \theta \le 1$ the denominator is at least one, so the implicit step is a bounded Fourier multiplier on ${\ell }^2$. Set $v={\sin }^2(\xi /2)\in [0,1]$ and use ${\sin }^2\xi =4v(1-v)$. The numerator is no greater than the denominator, so $G\le 1$. The other inequality isIf $\theta \ge 1/2$, the bracket is nonpositive and this always holds. If $0\le \theta <1/2$, the concave quadratic has its maximum at $v=(1-2\theta )/[2(1-\theta )]$, with value $(1-2\theta {)}^2/[4(1-\theta )]$. Therefore $|G|\le 1$ precisely under the stated conditionand without a mesh-ratio restriction when $1/2\le \theta \le 1$. The Parseval identity now givesIteration proves stability with constant one. It is conditional for $\theta <1/2$ and unconditional for $\theta \ge 1/2$. The sampled initial data have finite support, hence belong to ${\ell }^2$. This is the Fourier stability threshold for a mixed-width heat scheme.