Past exam of the numerical solutions of partial differential equations course of the University of Oxford / 2023 / 2

For $u^{(m,n)}(x,y)=\sin (m\pi x)\sin (n\pi y)$, direct differentiation yieldsSince the planar Laplacian is $\Delta ={\partial }_{xx}+{\partial }_{yy}$, the required eigenvalue of $-\Delta +I$ isOn the sides $x=0,1$ the first sine factor vanishes, and on $y=0,1$ the second sine factor vanishes. Thus the homogeneous Dirichlet boundary condition holds on the whole boundary. The eigenfunction is nonzero, for instance at $(x,y)=(1/(2m),1/(2n))$, where its value is one.

The five-point Laplacian on the square grid isIt adds the second-order central differences in the two coordinate directions and uses the given zero Dirichlet boundary conditions when a neighbouring node lies on the boundary.

The sine addition formula impliesApply this identity with $\alpha =m\pi h$ in the $x$ direction and with $\alpha =n\pi h$ in the $y$ direction. For $U_{i,j}^{(m,n)}=\sin (m\pi ih)\sin (n\pi jh)$, it givesHence the discrete eigenvalues areBoth sine factors vanish on their respective boundary sides. Also $U_{1,1}^{(m,n)}>0$, since $m\pi /N$ and $n\pi /N$ lie strictly between zero and $\pi$. Thus each array is a nonzero discrete eigenvector, rather than merely the zero solution of the homogeneous system. The separated structure is also described by the five-point Dirichlet Laplacian as a Kronecker sum.

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.