= Solution
Put $h=1/(m+1)$ and impose homogeneous <Dirichlet boundary conditions>, as required by the stated continuum <eigenvalues>. Let $T=\operatorname{tridiag}(1,-2,1)$ be the unscaled one-dimensional <Dirichlet discrete Laplacian>. For $1\leq k\leq m$, the vectors $v_j^{(k)}=\sin(jk\pi h)$ satisfy
$$
Tv^{(k)}=\tau_kv^{(k)},\qquad \tau_k=2\cos(k\pi h)-2=-4\sin^2\frac{k\pi h}{2}.
$$
The identity follows by adding the two neighboring <sines>; the values at $j=0,m+1$ vanish. These $m$ mutually orthogonal vectors form the <discrete sine transform> basis. Consequently their <tensor products> give all $m^2$ two-dimensional modes, not merely a few candidate <eigenvectors>.
In a compatible grid ordering the axial operator is $A_5=h^{-2}(T\otimes I+I\otimes T)$. The nine-point operator is
$$
A_9=A_5+\frac1{6h^2}T\otimes T.
$$
Indeed, the added <Kronecker product> supplies the diagonal neighbors with weight $1/6$, subtracts $1/3$ from each axial neighbor and adds $2/3$ to the central coefficient. Thus it has exactly the coefficients of the second discretization. On $v^{(k)}\otimes v^{(l)}$ the <eigenvalues> are
$$
\boxed{\lambda_{5,kl}=-\frac4{h^2}\left(\sin^2\frac{k\pi h}{2}+\sin^2\frac{l\pi h}{2}\right),}
$$
and
$$
\boxed{\lambda_{9,kl}=\lambda_{5,kl}+\frac8{3h^2}\sin^2\frac{k\pi h}{2}\sin^2\frac{l\pi h}{2},\qquad 1\leq k,l\leq m.}
$$
Equivalently, $h^2\lambda_{9,kl}=-10/3+4[\cos(k\pi h)+\cos(l\pi h)]/3+2\cos(k\pi h)\cos(l\pi h)/3$.
Back to article page
