Insert the Fourier mode
$$
u_m^n=\zeta^n e^{im\theta}
$$
into the <leapfrog advection scheme>. This gives
$$
\zeta^{n+1}
=\mu(e^{i\theta}-e^{-i\theta})\zeta^n+\zeta^{n-1},
$$
and hence the amplification polynomial
$$
\zeta^2-2i\mu\sin\theta\,\zeta-1=0.
$$
Its roots are
$$
\zeta_\pm=i\mu\sin\theta
\pm\sqrt{1-\mu^2\sin^2\theta}.
$$
If $0<\mu<1$, the square root is real for every $\theta$, and
$$
|\zeta_\pm|^2
=\mu^2\sin^2\theta+1-\mu^2\sin^2\theta=1.
$$
If $\mu>1$, choosing $|\sin\theta|>1/\mu$ makes the roots non-real multiples of $i$ with reciprocal moduli, one of which exceeds one, so the method is unstable.

The endpoint $\mu=1$ requires care. At $\theta=\pi/2$ the polynomial is
$$
(\zeta-i)^2,
$$
so a generic mode has the form $(c_1+c_2n)i^n$ and grows linearly. It satisfies the weaker test $|\zeta_\pm|\leq1$ often quoted in elementary Fourier analysis, but it violates the <root condition for a multistep method> because the unit-modulus root is repeated. Therefore the rigorous uniform stability range for arbitrary two-level initial data is
$$
\boxed{0<\mu<1}.
$$
If “stable” is being used only for the non-amplification condition on root moduli, the conventionally quoted range is $0<\mu\leq1$, with $\mu=1$ a defective marginal endpoint rather than a uniformly stable case.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 40E / c / Solution

Centered space and leapfrog time discretization of $u_t+cu_x=0$ gives
$$
u_j^{n+1}=u_j^{n-1}-\nu(u_{j+1}^n-u_{j-1}^n),
\qquad \nu=c\Delta t/\Delta x.
$$
Its amplification polynomial is $G^2+2i\nu\sin\theta\,G-1$. The two roots have unit modulus for $|\nu\sin\theta|\leq1$, but a repeated unit root at equality violates the uniform root condition.


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