Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 38C / b / Solution

With the endpoint values zero, move the new-level differences to the left and keep the old-level differences on the right. Then $B$ is tridiagonal with diagonal $1+\mu$ and off-diagonal entries $-\mu /2$, while $C$ has diagonal $1-\mu$ and off-diagonal entries $\mu /2$. Thus $B{u}^{n+1}=C{u}^n$.

Part (a) supplies their common orthonormal eigenbasis. Their eigenvalues are $1+\mu (1-\cos {\theta }_k)$ and $1-\mu (1-\cos {\theta }_k)$. For $\mu \ge 0$, $B$ is invertible, and the step matrix $B^{-1}C$ has eigenvaluesSince the eigenbasis is orthonormal, $\Vert (B^{-1}C{)}^n{\Vert }_2={\max }_k|{\eta }_k{|}^n\le 1$, uniformly in the number of steps and mesh. Hence the method is unconditionally stable for every $\mu \ge 0$, which includes all physical positive time steps. For negative $\mu$, every nonzero modal parameter has modulus larger than one whenever its denominator is nonzero; if a denominator vanishes the step is undefined. Thus the algebraic stable range is exactly $[0,\infty )$. This proof uses simultaneous matrix diagonalization, not Fourier stability analysis.