 Solution

Put $\nu =k/\delta$ and use the discrete Fourier transform on the infinite grid. A spatial Fourier mode $e^{i(j\theta +\ell \varphi )}$ has time coefficient satisfyingIts amplification polynomial of a multilevel finite difference scheme is ${\zeta }^2-2is\zeta -1$, with rootsFor $|s|<1$ they are distinct and have modulus one. Since $|s|\le 2\nu$, 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 thereforeThe 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.