Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 72 / 5

Interpret the square as $[0,1{]}^2$, use equal spacing $h=1/J$ in both directions and retain interior indices $1\le m,n\le J-1$. Boundary values contribute to the right-hand side and do not change the interior coefficient matrix. Products of the one-dimensional sine modes used in the discrete sine transformform a basis, and substitution into the five-point Laplacian givesThe interior matrix is ${\Delta }_h+\lambda I$, so it is nonsingular exactly whenThese are the discrete Helmholtz resonance values, including their possible multiplicities. A sufficient condition is $\lambda <{\Lambda }_{11}=8h^{-2}{\sin }^2(\pi /(2J))$, so in particular every nonpositive $\lambda$ is allowed. That is not a necessary restriction: positive values between, or above, the finitely many discrete eigenvalues are nonsingular as well. With homogeneous Dirichlet boundary conditions a nonresonant system has only the zero solution; at a resonance the corresponding modes of the discrete sine transform give nonzero homogeneous solutions.

For a sufficiently smooth function, Taylor expansion of the four neighboring values givesThe reaction term $\lambda u$ is sampled exactly at the grid point. On an exact solution, therefore,The method has second-order local accuracy for the partial differential equation. The original printed equations multiply this expression by $h^2$, so their unscaled residual is $O(h^4)$; that scaling does not make the Laplacian approximation fourth order. Resonance and inverse-matrix bounds are separate issues from this local truncation calculation.

Let $S_{1}u$ be the sum of the four axial neighbors and $S_{2}u$ the sum of the four diagonal neighbors. The two PDF stencils defineThe reaction stencil has four axial weights $1/12$ and center weight $2/3$, so its weights sum to one. Taylor expansion givesFor example, the diagonal neighbors supply the $u_{xxyy}$ term needed to turn the fourth derivatives into the full squared Laplacian. Adding the reaction contribution producesOn a smooth solution of the constant-coefficient equation, both displayed leading terms vanish. Thus the compact fourth-order Helmholtz stencil hasThe cancellation uses the differential equation and constant $\lambda$; the nine-point finite-difference stencil alone still has an $O(h^2)$ operator error on an arbitrary smooth function. It also does not establish a uniform global fourth-order error without an appropriate inverse bound and compatible treatment of boundaries and resonances.