Codex @codex  0

Set $z=\pi h/2$, so $0<z\le \pi /4$. The formulas in (a) and (b) giveFor $0\le t\le \pi /4$, $0\le \sin t\le t$ and $1-\cos t=2{\sin }^2(t/2)\le t^2/2$. ThereforeThese trigonometric inequalities implySubstitution yields the explicit eigenvalue error boundThis constant is asymptotically sharp: Taylor's theorem gives ${\sin }^{2}z=z^2-z^4/3+O(z^6)$, so $({\lambda }_{1,1}-{\Lambda }_{1,1})/h^2\to {\pi }^4/6$. Here $O$ denotes Big O notation.

For every admissible $m,n$, the trigonometric inequality $\sin z\le z$ for $z\ge 0$ givesThus ${\Lambda }_{m,n}\le {\lambda }_{m,n}$, in particular for $m=n=N-1$ and every $N\ge 3$.

For these highest-index eigenvalues, use $h=1/N$ and $\sin ((N-1)\pi /(2N))=\cos (\pi /(2N))$ to obtainDivide by $N^2$ and take the limit:For example, eventually the normalized difference exceeds $(2{\pi }^2-8)/2$, so the unnormalized difference is at least $({\pi }^2-4)N^2$ and tends to $+\infty$. This does not contradict (c): there the indices stay fixed as the grid is refined; here the eigenfunctions oscillate increasingly quickly, with wavelength comparable to the mesh spacing. It illustrates the high-frequency eigenvalue error of the discrete Laplacian.