= Solution
<Fourier transform> on the infinite spatial lattice gives
$$
a_{n+1}=2iA(\theta)a_n+a_{n-1},\qquad
A(\theta)=a_1\sin\theta+a_2\sin2\theta,
$$
with <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 give
$$
A(\theta)=\frac58\sin\theta-\frac1{16}\sin2\theta
=\frac{\sin\theta(5-\cos\theta)}8,\qquad |A(\theta)|\le\frac34<1.
$$
Consequently \b[$\mu=1/2$ is stable].
At $\mu=3/2$,
$$
A(\theta)=\frac78\sin\theta+\frac5{16}\sin2\theta,
\qquad A(\pi/3)=\frac{19\sqrt3}{32}>1.
$$
The growing <polynomial root> at this phase has modulus
$$
\boxed{|G|=\frac{19\sqrt3+\sqrt{59}}{32}>1.}
$$
To 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 $\|u^n\|_2\ge(1+\delta)^n\|u^0\|_2$, 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 $T$, \b[$\mu=3/2$ is unstable].
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