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For Neumann boundaries the correct eigenfunctions include zero indices:Their squared wave number is $q=(n_x\pi /L_x{)}^2+(n_y\pi /L_y{)}^2$. The constant pair $(0,0)$ is reaction-stable. Instability requires $q_{-}<q<q_{+}$, using the band derived above. If both side lengths are large, the wave-number spacings are small enough that admissible modes can enter the band in either direction, including mixed modes. This does not mean the single $(1,1)$ mode necessarily stays unstable as lengths tend to infinity: its wave number eventually tends to zero below the band. Rather, increasingly fine wave-number spacing permits some suitable higher integers.

Thin-domain Turing mode selection applies if $L_y<\pi /\sqrt{q_{+}}$, every mode with $n_y\ge 1$ has $q\ge (\pi /L_y{)}^2>q_{+}$ and is stable. A sufficiently large $L_x$ can nevertheless admit an integer $n_x\ge 1$ with $q_{-}<(n_x\pi /L_x{)}^2<q_{+}$ and $n_y=0$. HenceThe printed range starting both indices at one excludes these purely longitudinal modes and cannot imply the requested conclusion. It also omits the constant mode. Further, each allowed spatial eigenfunction generally has two temporal eigenvalue branches of $J-q\mathcal{D}$, with coefficients along their eigenvectors; a single unrestricted scalar $\lambda (q)$ is shorthand for those branches rather than a complete general expansion.