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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.

Use a spatial discrete Fourier mode $u_{k,j}^n=a_n{e}^{i(k\xi +j\eta )}$. Its recurrence and amplification polynomial of a multilevel finite difference scheme areThe polynomial roots are $G_{\pm }=is\pm \sqrt{1-s^2}$. For $|s|<1$ both have modulus one and are separated. Since $|\sin \xi +\sin \eta |\le 2$, a fixed $0<\mu <1/2$ bounds their separation below by $2\sqrt{1-4{\mu }^2}$. The two-level companion matrix is therefore diagonalizable with uniformly bounded eigenvector matrix and inverse. To see the uniform bound explicitly, write $V=\left(\begin{array}{cc}{G}_{+}& G_{-}\\ 1& 1\end{array}\right)$. Its determinant is $G_{+}-G_{-}$; all its entries have modulus one, and those of $V^{-1}$ are bounded by the reciprocal of the root separation. The companion matrix powers are $V\mathrm{diag}(G_{+}^n,G_{-}^n)V^{-1}$, bounded independently of frequency and step number. Parseval's identity converts this into a mesh-independent discrete $L^2$ bound for arbitrary perturbations at the two initial levels.

If $\mu >1/2$, choose $\xi =\eta =\pi /2$. Then $s=2\mu >1$, and one polynomial root has modulus $s+\sqrt{s^2-1}>1$. On periodic meshes with the number of points divisible by four, this is an actual grid mode, producing exponential linear instability.

At $\mu =1/2$ the same phases give $(G-i{)}^2$. The companion matrix is not a scalar matrix, so the double polynomial root has a nontrivial Jordan block. The solution $a_n=n{i}^n$ has bounded starting amplitudes but grows like $n$. On a fixed physical time interval $n$ is of order $1/k$, so no mesh-independent stability bound exists. Consequentlyis the stability interval under the standard two-level definition. Testing only polynomial root moduli would misleadingly include the endpoint. A special startup selecting the non-growing branch at that endpoint restricts the perturbations and does not establish stability of the full recurrence. This is the two-dimensional leapfrog stability threshold.