For a <discrete Fourier mode> $u_m^n=r^ne^{im\theta}$, put
$$
q=4\mu\sin^2\frac\theta2,
\qquad 0\leq q\leq4\mu.
$$
The recurrence becomes
$$
r^{n+1}=\left(1-\frac32q\right)r^n+\frac12qr^{n-1},
$$
so its <amplification polynomial of a multilevel finite difference scheme> is
$$
p(r)=r^2-\left(1-\frac32q\right)r-\frac12q.
$$
Its roots are real. Moreover,
$$
p(1)=q,
\qquad
p(-1)=2(1-q),
\qquad
\frac{r_1+r_2}{2}=\frac12-\frac34q.
$$
The root-location criterion in the question therefore puts both roots in $[-1,1]$ exactly when $0\leq q\leq1$. <Fourier stability analysis> for every <wavenumber> is equivalent to $4\mu\leq1$, so
$$
\boxed{\text{the scheme is stable if and only if }\mu\leq\frac14.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 41E / b / Solution

Fourier stability analysis transforms a translation-invariant finite-difference scheme into scalar or matrix recurrences indexed by wavenumber. The <Parseval identity> then converts uniform multiplier bounds into discrete $2$-norm bounds.

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