A four-node cyclic chain mode with $a=c=1$: exact ODE check, dispersion value ${\omega }^2=2$, and the continuum error bound

Dispersion: $\frac{4c^2}{a^2}{\sin }^2(\kappa a\mathrm{/}2)=4{\sin }^2(\pi \mathrm{/}4)=4\cdot \frac{1}{2}=2$, so by [L1] the ansatz with $\omega =\sqrt{2}$ solves the chain equation; explicitly, $q_j^{\mathrm{\prime }\mathrm{\prime }}=-2q_j$ and $(q_{j+1}-2q_j+q_{j-1})=-2q_j$ by [L2], matching the equation $q_j^{\mathrm{\prime }\mathrm{\prime }}=(q_{j+1}-2q_j+q_{j-1})$.

Error bound: put $z:=\kappa a\mathrm{/}2=\pi \mathrm{/}4$. By [L3], twice integrating $\mathrm{\mid }\sin r\mathrm{\mid }\le \mathrm{\mid }r\mathrm{\mid }$ gives $\mathrm{\mid }\cos s-1\mathrm{\mid }\le s^2\mathrm{/}2$ and $\mathrm{\mid }\sin z-z\mathrm{\mid }\le \mathrm{\mid }z{\mathrm{\mid }}^3\mathrm{/}6$; also $\mathrm{\mid }\sin z+z\mathrm{\mid }\le 2\mathrm{\mid }z\mathrm{\mid }$. Hence $\mathrm{\mid }4{\sin }^{2}z-4z^2\mathrm{\mid }\le 4\mathrm{\mid }\sin z-z\mathrm{\mid }\mathrm{\mid }\sin z+z\mathrm{\mid }\le 4\mathrm{\mid }z{\mathrm{\mid }}^4\mathrm{/}3$, so $\mathrm{\mid }{\omega }^2-c^2{\kappa }^2\mathrm{\mid }\le c^2{\kappa }^4{a}^2\mathrm{/}12={\pi }^4\mathrm{/}192$ directly at this compatible mode. The exact error is $\mathrm{\mid }2-{\pi }^2\mathrm{/}4\mathrm{\mid }={\pi }^2\mathrm{/}4-2$, as follows without a numerical bound on $\pi$: for $z=\pi \mathrm{/}4>0$, $\mathrm{\mid }\cos s\mathrm{\mid }\le 1$ on $[0,z]$ and $\mathrm{\mid }\cos z\mathrm{\mid }=1\mathrm{/}\sqrt{2}<1$. Continuity gives a nonempty terminal interval where $\mathrm{\mid }\cos s\mathrm{\mid }<1$, so The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$ gives $\mathrm{\mid }\sin z\mathrm{\mid }=\mathrm{\mid }{\int }_0^z\cos sds\mathrm{\mid }<z$. Thus $4z^2>4{\sin }^{2}z=2$. Moreover $\mathrm{\mid }\sin z+z\mathrm{\mid }<2z$, so the same remainder product gives a strictly smaller error than ${\pi }^4\mathrm{/}192$. This proves the exact comparison without applying the supplier's restricted $\mathrm{\mid }\kappa a\mathrm{\mid }\le 1$ clause at $\kappa a=\pi \mathrm{/}2$.

The example verifies the exact discrete dispersion at explicit parameters and checks that the continuum error bound is satisfied, not saturated, at this wave number.