 Solution

Use the Fourier transform for the spatial Cauchy problem, or its periodic analogue, and regard the two starting levels as independently perturbed data. A Fourier mode with spatial factor $e^{im\theta }$ has amplification roots $G$ satisfyingPut $c=2\mu -1$ and $G=e^{i\theta /2}q$. ThenIf $|c|<1$, both roots have modulus one and their separation is bounded below uniformly in frequency:The uniform power bound from separated amplification roots now controls the two-level companion matrix for every time step. Its entries are uniformly bounded, and its eigenvector conditioning is bounded by the reciprocal root gap. The Parseval identity transfers this frequency-uniform bound to the spatial ${\ell }^2$ norm. This proves stability, rather than merely checking each root's modulus.

If $|c|>1$, the frequency $\theta =\pi$ has a root outside the unit disk, so there is exponential instability. If $|c|=1$, the two roots at $\theta =\pi$ coincide on the unit circle. The companion matrix is not a scalar matrix and has a nontrivial Jordan block; its powers grow linearly in the number of steps. Frequencies arbitrarily near that value produce the same lack of a uniform bound for localized Fourier packets, so this also invalidates Cauchy ${\ell }^2$ stability, not only periodic plane-wave stability. At $\mu =1$ the double amplification root is $-1$, and at $\mu =0$ it is $1$.

Therefore the full two-level stability range for a fixed positive Courant ratio isThe endpoint $\mu =0$ is moreover not a positive time step. Bounds deteriorate as $\mu$ approaches either endpoint; the displayed range is not a uniform claim over ratios arbitrarily close to one. A prescribed starter that removes one special parasitic component can change behavior for selected initial data, but does not establish the requested unrestricted two-level stability.