 Solution

Fourier transform on the infinite spatial lattice giveswith polynomial roots $G_{\pm }=iA\pm \sqrt{1-A^2}$. When $|A|$ is uniformly less than one, the polynomial roots have modulus one and a uniform positive separation. Diagonalizing the companion matrix then bounds all its powers independently of $\theta$ and $n$. Parseval's identity proves stability in the discrete $L^2$ norm for arbitrary square-summable starting perturbations at both levels.

At $\mu =1/2$ the coefficients giveConsequently $\mu =1/2$ is stable.

At $\mu =3/2$,The growing polynomial root at this phase has modulusTo turn this into a Cauchy problem linear instability proof, rather than rely on a plane wave of infinite norm, choose a small interval around $\pi /3$ on which the growing polynomial root has modulus at least $1+\delta$. Choose an initial Fourier transform amplitude that is a nonzero square-integrable function supported there, and set the level-one amplitude equal to that polynomial root times the level-zero amplitude. The recurrence then multiplies by the same growing polynomial root at every step on the support. Parseval identity gives $\Vert u^n{\Vert }_2\ge (1+\delta {)}^n\Vert u^0{\Vert }_2$, whereas both initial norms are bounded independently of $h$. Real data can be obtained by adding the conjugate interval around $-\pi /3$. Since $n=T/(\mu h)$ tends to infinity at fixed $T$, $\mu =3/2$ is unstable.