= Solution
Put $\nu=k/\delta$ and use the <discrete Fourier transform> on the infinite grid. A spatial <Fourier mode> $e^{i(j\theta+\ell\phi)}$ has time coefficient satisfying
$$
v_{n+1}=2is\,v_n+v_{n-1},\qquad s=\nu(\sin\theta+\sin\phi).
$$
Its <amplification polynomial of a multilevel finite difference scheme> is $\zeta^2-2is\zeta-1$, with roots
$$
\zeta_\pm=is\pm\sqrt{1-s^2}.
$$
For $|s|<1$ they are distinct and have modulus one. Since $|s|\leq2\nu$, a fixed $0<\nu<1/2$ bounds their separation below by $2\sqrt{1-4\nu^2}$. Solving for the two mode amplitudes therefore bounds every power of the two-level <companion matrix> uniformly in frequency and step number. <Parseval identity> transfers this bound to the <discrete L2 norm>.
If $\nu>1/2$, an open set of frequencies around $(\pi/2,\pi/2)$ has $|s|>1$ and a root of modulus greater than one. Fourier data supported there grow exponentially. At $\nu=1/2$, the central frequency has the double root $i$, and the recurrence admits $v_n=(A+Bn)i^n$. Thus the root moduli being one do not suffice: a nontrivial <Jordan block> makes the two-level powers grow linearly. Although this exact frequency has measure zero on the infinite grid, the <matrices> depend continuously on frequency. For every step count, a neighborhood has powers nearly as large, so their essential supremum is still unbounded. Wave packets in those neighborhoods rule out a uniform <stability of a numerical method> estimate in the <discrete L2 norm>.
For arbitrary perturbations in both starting levels the sharp positive range is therefore
$$
\boxed{0<\Delta t/\Delta x<1/2.}
$$
The zero-step limiting recurrence is bounded but does not advance time. A specially filtered startup can suppress a repeated-root mode; that restriction does not establish stability of the full two-level scheme. This is the <two-dimensional leapfrog stability threshold>.
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