Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 40E (Numerical Analysis)

Termwise differentiation of a Fourier series multiplies a mode by its wavenumber, soThe convolution theorem turns pointwise multiplication into discrete convolution:soThe assumed real analyticity justifies these termwise operations through rapid decay of the Fourier coefficients.

Expand $a,u,f$ in the two-dimensional Fourier basis. Applying the derivative and product rules from part (a), the $(m,n)$ coefficient of the equation isThe normalization and compatibility conditions areThus, on ${\mathbb{Z}}^2\setminus \{(0,0)\}$, the infinite Fourier spectral method system isFor a square spectral truncation, chooseretain equations and unknowns indexed by ${\Lambda }_N$, and set omitted coefficients to zero. This is the Fourier–Galerkin method: it projects the residual onto the retained Fourier modes. Other finite mode sets may be used in the same way.

Here the only nonzero coefficients of $a$ areTherefore, for each $(m,n)\in {\Lambda }_N$, the truncated system iswhere a coefficient is omitted when its index is outside ${\Lambda }_N$ or equals $(0,0)$.

For wavevectors $\mathbf{k}=(m,n)$ and $\mathbf{\ell }=(p,q)$, the matrix entry isBecause ${\hat{a}}_{\mathbf{r}}={\hat{a}}_{-\mathbf{r}}$, $A_{\mathbf{k},\mathbf{\ell }}=A_{\mathbf{\ell },\mathbf{k}}$, so the matrix is symmetric. Its diagonal entry is $5{\pi }^2|\mathbf{k}{|}^2$. The sum of absolute off-diagonal entries is at mostbecause $|\mathbf{k}{|}^2\pm k_j\ge 0$ for nonzero integer $\mathbf{k}$; truncation can only reduce the sum. Hence $A$ is strictly diagonally dominant with positive diagonal. The Gershgorin circle theorem places every eigenvalue in the positive intervalfor at least one row. Since the symmetric matrix has only real eigenvalues, all are positive, and $A$ is a positive-definite matrix. Any one-dimensional ordering merely conjugates $A$ by a permutation matrix, preserving symmetry and positive definiteness.